TÍNH GIỚI HẠN DẠNG VÔ ĐỊNH \(\frac{\infty }{\infty }\)
Phương pháp:
L = \(\mathop {\lim }\limits_{x \to \pm \infty } \frac{{P(x)}}{{Q(x)}}\) trong đó$P(x),Q(x) \to \infty $, dạng này ta còn gọi là dạng vô định$\frac{\infty }{\infty }$.
với P(x), Q(x) là các đa thức hoặc các biểu thức chứa căn.
Phương pháp:
L = \(\mathop {\lim }\limits_{x \to \pm \infty } \frac{{P(x)}}{{Q(x)}}\) trong đó$P(x),Q(x) \to \infty $, dạng này ta còn gọi là dạng vô định$\frac{\infty }{\infty }$.
với P(x), Q(x) là các đa thức hoặc các biểu thức chứa căn.
- Nếu P(x), Q(x) là các đa thức thì chia cả tử và mẫu cho luỹ thừa cao nhất của x.
- Nếu P(x), Q(x) có chứa căn thì có thể chia cả tử và mẫu cho luỹ thừa cao nhất của x hoặc nhân lượng liên hợp.
- $\mathop {\lim }\limits_{x \to + \infty } {x^{2k}} = + \infty ;\,\mathop {\lim }\limits_{x \to - \infty } {x^{2k}} = + \infty ;\mathop {\lim }\limits_{x \to + \infty } {x^{2k + 1}} = + \infty ;\,\mathop {\lim }\limits_{x \to - \infty } {x^{2k + 1}} = + \infty $
- $\mathop {\lim }\limits_{\begin{array}{*{20}{c}} {{\rm{x}} \to + \infty }\\ {(x \to - \infty )} \end{array}} \frac{k}{{{x^n}}} = 0\,\,(n > 0;k \ne 0)$
- $\mathop {\lim }\limits_{x \to {x_0}} f(x) = + \infty {\rm{ }}( - \infty ) \Leftrightarrow \mathop {\lim }\limits_{x \to {x_0}} \frac{k}{{f(x)}} = 0{\rm{ }}(k \ne 0)$.
Câu 1. \(\mathop {\lim }\limits_{x \to \infty } \frac{5}{{3x + 2}}\) bằng:
A. 0.
B. 1.
C. \(\frac{5}{3}\).
D. + ∞.
Chọn A
Cách 1: \(\mathop {\lim }\limits_{x \to \infty } \frac{5}{{3x + 2}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{5}{x}}}{{3 + \frac{2}{x}}} = 0\)
Cách 2: Bấm máy tính như sau: \(\frac{5}{{3x + 2}}\) + CACL + \(x = {10^9}\)và so đáp án (với máy casio 570 VN Plus)
Cách 3: Dùng chức lim của máy VNCALL 570ES Plus: ${\left. {\lim \frac{5}{{3x + 2}}} \right|_{x \to {{10}^9}}}$ và so đáp án.
Cách 1: \(\mathop {\lim }\limits_{x \to \infty } \frac{5}{{3x + 2}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{5}{x}}}{{3 + \frac{2}{x}}} = 0\)
Cách 2: Bấm máy tính như sau: \(\frac{5}{{3x + 2}}\) + CACL + \(x = {10^9}\)và so đáp án (với máy casio 570 VN Plus)
Cách 3: Dùng chức lim của máy VNCALL 570ES Plus: ${\left. {\lim \frac{5}{{3x + 2}}} \right|_{x \to {{10}^9}}}$ và so đáp án.
A. - 1
B. 1.
C. 7.
D. + ∞
Chọn B
$\mathop {\lim }\limits_{x \to + \infty } \frac{{{x^4} + 7}}{{{x^4} + 1}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{1 + \frac{7}{{{x^4}}}}}{{1 + \frac{1}{{{x^4}}}}} = 1$.
$\mathop {\lim }\limits_{x \to + \infty } \frac{{{x^4} + 7}}{{{x^4} + 1}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{1 + \frac{7}{{{x^4}}}}}{{1 + \frac{1}{{{x^4}}}}} = 1$.
A. + ∞
B. - ∞
C. \(\frac{{2 - \sqrt 3 }}{6}\)
D. 0
Ta có: \(C = \mathop {\lim }\limits_{x \to + \infty } \frac{{2 - \sqrt {3 + \frac{2}{{{x^2}}}} }}{{5 + \sqrt {1 + \frac{1}{{{x^2}}}} }} = \frac{{2 - \sqrt 3 }}{6}\)
A. - 2.
B. - 1/3.
C. 1/3.
D. 2.
Chọn D
Cách 1: $\mathop {\lim }\limits_{x \to \infty } \frac{{2{x^2} - 1}}{{3 - {x^2}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{2 - \frac{1}{{{x^2}}}}}{{\frac{3}{{{x^2}}} - 1}} = 2$
Cách 2: Bấm máy tính như sau: \(\frac{{2{x^2} - 1}}{{3 - {x^2}}}\) + CACL + \(x = {10^9}\) và so đáp án.
Cách 3: Dùng chức lim của máy VNCALL 570ES Plus: ${\left. {\lim \frac{{2{x^2} - 1}}{{3 - {x^2}}}} \right|_{x \to {{10}^9}}}$ và so đáp án.
Cách 1: $\mathop {\lim }\limits_{x \to \infty } \frac{{2{x^2} - 1}}{{3 - {x^2}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{2 - \frac{1}{{{x^2}}}}}{{\frac{3}{{{x^2}}} - 1}} = 2$
Cách 2: Bấm máy tính như sau: \(\frac{{2{x^2} - 1}}{{3 - {x^2}}}\) + CACL + \(x = {10^9}\) và so đáp án.
Cách 3: Dùng chức lim của máy VNCALL 570ES Plus: ${\left. {\lim \frac{{2{x^2} - 1}}{{3 - {x^2}}}} \right|_{x \to {{10}^9}}}$ và so đáp án.
A. 1/2.
B. \(\frac{{\sqrt 2 }}{2}\).
C. 0.
D. + ∞.
Chọn C
Cách 1: \(\mathop {\lim }\limits_{x \to + \infty } \sqrt {\frac{{{x^2} + 1}}{{2{x^4} + {x^2} - 3}}} \)\( = \mathop {\lim }\limits_{x \to + \infty } \sqrt {\frac{{\frac{1}{{{x^2}}} + \frac{1}{{{x^4}}}}}{{2 + \frac{1}{{{x^2}}} - \frac{3}{{{x^4}}}}}} = 0\)
Cách 2: Bấm máy tính như sau: $\frac{5}{{3x + 2}}$ + CACL + \(x = {10^9}\) và so đáp án.
Cách 3: Dùng chức lim của máy VNCALL 570ES Plus: ${\left. {\lim \sqrt {\frac{{{x^2} + 1}}{{2{x^4} + {x^2} - 3}}} } \right|_{x \to {{10}^9}}}$ và so đáp án.
Cách 1: \(\mathop {\lim }\limits_{x \to + \infty } \sqrt {\frac{{{x^2} + 1}}{{2{x^4} + {x^2} - 3}}} \)\( = \mathop {\lim }\limits_{x \to + \infty } \sqrt {\frac{{\frac{1}{{{x^2}}} + \frac{1}{{{x^4}}}}}{{2 + \frac{1}{{{x^2}}} - \frac{3}{{{x^4}}}}}} = 0\)
Cách 2: Bấm máy tính như sau: $\frac{5}{{3x + 2}}$ + CACL + \(x = {10^9}\) và so đáp án.
Cách 3: Dùng chức lim của máy VNCALL 570ES Plus: ${\left. {\lim \sqrt {\frac{{{x^2} + 1}}{{2{x^4} + {x^2} - 3}}} } \right|_{x \to {{10}^9}}}$ và so đáp án.
A. \( - \frac{{3\sqrt 2 }}{2}\).
B. \(\frac{{\sqrt 2 }}{2}\).
C. \(\frac{{3\sqrt 2 }}{2}\).
D. \( - \frac{{\sqrt 2 }}{2}\).
Chọn A
Cách 1: \(\mathop {\lim }\limits_{x \to - \infty } \frac{{1 + 3x}}{{\sqrt {2{x^2} + 3} }} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{1}{{{x^2}}} + 3}}{{ - \sqrt {2 + \frac{3}{{{x^2}}}} }} = - \frac{{3\sqrt 2 }}{2}\)
Cách 2: Bấm máy tính như sau: $\frac{{2{x^2} - 1}}{{3 - {x^2}}}$ + CACL + \(x = - {10^9}\) và so đáp án.
Cách 3: Dùng chức lim của máy VNCALL 570ES Plus: ${\left. {\lim \frac{{1 + 3x}}{{\sqrt {2{x^2} + 3} }}} \right|_{x \to - {{10}^9}}}$ và so đáp án.
Cách 1: \(\mathop {\lim }\limits_{x \to - \infty } \frac{{1 + 3x}}{{\sqrt {2{x^2} + 3} }} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{1}{{{x^2}}} + 3}}{{ - \sqrt {2 + \frac{3}{{{x^2}}}} }} = - \frac{{3\sqrt 2 }}{2}\)
Cách 2: Bấm máy tính như sau: $\frac{{2{x^2} - 1}}{{3 - {x^2}}}$ + CACL + \(x = - {10^9}\) và so đáp án.
Cách 3: Dùng chức lim của máy VNCALL 570ES Plus: ${\left. {\lim \frac{{1 + 3x}}{{\sqrt {2{x^2} + 3} }}} \right|_{x \to - {{10}^9}}}$ và so đáp án.
A. + ∞
B. - ∞
C. \(\frac{4}{3}\)
D. 1
Ta có: \(D = \mathop {\lim }\limits_{x \to - \infty } \frac{{{x^2}\sqrt[3]{{\frac{1}{{{x^6}}} + \frac{1}{{{x^2}}} + 1}}}}{{{x^2}\sqrt {\frac{1}{{{x^4}}} + \frac{1}{{{x^2}}} + 1} }} = 1\)
A. 0.
B. 1/2.
C. 1.
D. Không tồn tại.
Chọn A
\(\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to + \infty } \left( {x + 2} \right)\sqrt {\frac{{x - 1}}{{{x^4} + {x^2} + 1}}} = \mathop {\lim }\limits_{x \to + \infty } \sqrt {\frac{{\left( {x - 1} \right)\left( {x + 2} \right)}}{{{x^4} + {x^2} + 1}}} = \mathop {\lim }\limits_{x \to + \infty } \sqrt {\frac{{\frac{1}{{{x^2}}} + \frac{1}{{{x^3}}} - \frac{2}{{{x^4}}}}}{{1 + \frac{1}{{{x^2}}} + \frac{1}{{{x^4}}}}}} = 0\).
\(\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to + \infty } \left( {x + 2} \right)\sqrt {\frac{{x - 1}}{{{x^4} + {x^2} + 1}}} = \mathop {\lim }\limits_{x \to + \infty } \sqrt {\frac{{\left( {x - 1} \right)\left( {x + 2} \right)}}{{{x^4} + {x^2} + 1}}} = \mathop {\lim }\limits_{x \to + \infty } \sqrt {\frac{{\frac{1}{{{x^2}}} + \frac{1}{{{x^3}}} - \frac{2}{{{x^4}}}}}{{1 + \frac{1}{{{x^2}}} + \frac{1}{{{x^4}}}}}} = 0\).
A. \(3\).
B. 1/2.
C. 1.
D. + ∞.
Chọn A
$\mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {{x^2} - x + 3} }}{{2\left| x \right| - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{x\sqrt {1 - \frac{1}{x} + \frac{3}{{{x^2}}}} }}{{2x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{x\sqrt {1 - \frac{1}{x} + \frac{3}{{{x^2}}}} }}{{x\left( {2 - \frac{1}{x}} \right)}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {1 - \frac{1}{x} + \frac{3}{{{x^2}}}} }}{{\left( {2 - \frac{1}{x}} \right)}} = 3.$.
$\mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {{x^2} - x + 3} }}{{2\left| x \right| - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{x\sqrt {1 - \frac{1}{x} + \frac{3}{{{x^2}}}} }}{{2x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{x\sqrt {1 - \frac{1}{x} + \frac{3}{{{x^2}}}} }}{{x\left( {2 - \frac{1}{x}} \right)}} = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {1 - \frac{1}{x} + \frac{3}{{{x^2}}}} }}{{\left( {2 - \frac{1}{x}} \right)}} = 3.$.
A. \( - \frac{{21}}{5}\).
B. \(\frac{{21}}{5}\).
C. \( - \frac{{24}}{5}\).
D. \(\frac{{24}}{5}\).
Chọn C
$\mathop {\lim }\limits_{x \to + \infty } \frac{{{x^4} + 8x}}{{{x^3} + 2{x^2} + x + 2}}$ thành$\mathop {\lim }\limits_{x \to - 2} \frac{{{x^4} + 8x}}{{{x^3} + 2{x^2} + x + 2}}$
$\mathop {\lim }\limits_{x \to - 2} \frac{{{x^4} + 8x}}{{{x^3} + 2{x^2} + x + 2}} = \mathop {\lim }\limits_{x \to - 2} \frac{{x\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}}{{\left( {x + 2} \right)\left( {{x^2} + 1} \right)}} = \mathop {\lim }\limits_{x \to - 2} \frac{{x\left( {{x^2} - 2x + 4} \right)}}{{\left( {{x^2} + 1} \right)}} = - \frac{{24}}{5}.$
$\mathop {\lim }\limits_{x \to + \infty } \frac{{{x^4} + 8x}}{{{x^3} + 2{x^2} + x + 2}}$ thành$\mathop {\lim }\limits_{x \to - 2} \frac{{{x^4} + 8x}}{{{x^3} + 2{x^2} + x + 2}}$
$\mathop {\lim }\limits_{x \to - 2} \frac{{{x^4} + 8x}}{{{x^3} + 2{x^2} + x + 2}} = \mathop {\lim }\limits_{x \to - 2} \frac{{x\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}}{{\left( {x + 2} \right)\left( {{x^2} + 1} \right)}} = \mathop {\lim }\limits_{x \to - 2} \frac{{x\left( {{x^2} - 2x + 4} \right)}}{{\left( {{x^2} + 1} \right)}} = - \frac{{24}}{5}.$
A. + ∞
B. - ∞
C. - 1/2
D. 0
Ta có: \(E = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - x + 1}}{{\sqrt {{x^2} - x + 1} + x}} = - \frac{1}{2}\)
A. + ∞
B. - ∞
C. \(\frac{4}{3}\)
D. 0
Ta có: \(F = \mathop {\lim }\limits_{x \to - \infty } {x^2}\left( { - \sqrt {4 + \frac{1}{{{x^2}}}} - 1} \right) = - \infty \)
A. - ∞.
B. 0.
C. 4.
D. + ∞.
Chọn A
$\mathop {\lim }\limits_{x \to - \infty } \left( {4{x^5} - 3{x^3} + x + 1} \right) = \mathop {\lim }\limits_{x \to - \infty } {x^5}\left( {4 - \frac{3}{{{x^2}}} + \frac{1}{{{x^4}}} + \frac{1}{{{x^5}}}} \right) = - \infty .$.
$\mathop {\lim }\limits_{x \to - \infty } \left( {4{x^5} - 3{x^3} + x + 1} \right) = \mathop {\lim }\limits_{x \to - \infty } {x^5}\left( {4 - \frac{3}{{{x^2}}} + \frac{1}{{{x^4}}} + \frac{1}{{{x^5}}}} \right) = - \infty .$.
A. - ∞.
B. 0.
C. 1.
D. + ∞.
Chọn D
$\mathop {\lim }\limits_{x \to + \infty } \sqrt {{x^4} - {x^3} + {x^2} - x} = \mathop {\lim }\limits_{x \to + \infty } \sqrt {{x^4}\left( {1 - \frac{1}{x} + \frac{1}{{{x^2}}} - \frac{1}{{{x^3}}}} \right)} = + \infty .$.
$\mathop {\lim }\limits_{x \to + \infty } \sqrt {{x^4} - {x^3} + {x^2} - x} = \mathop {\lim }\limits_{x \to + \infty } \sqrt {{x^4}\left( {1 - \frac{1}{x} + \frac{1}{{{x^2}}} - \frac{1}{{{x^3}}}} \right)} = + \infty .$.
A. + ∞
B. - ∞
C. \(\frac{4}{3}\)
D. 0
Ta có: \(B = \mathop {\lim }\limits_{x \to - \infty } \left( {x - \left| x \right|\sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} } \right) = \mathop {\lim }\limits_{x \to - \infty } x\left( {1 + \sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} } \right) = - \infty \)
A. + ∞
B. - ∞
C. \(\frac{4}{3}\)
D. Đáp án khác
Ta có: \(M = \mathop {\lim }\limits_{x \to \pm \infty } \frac{{4x}}{{\sqrt {{x^2} + 3x + 1} + \sqrt {{x^2} - x + 1} }} = \left\{ \begin{array}{l}2{\rm{ khi }}x \to + \infty \\ - 2{\rm{ khi }}x \to - \infty \end{array} \right.\)
A. + ∞
B. - ∞
C. \(\frac{4}{3}\)
D. 0
Ta có: \(N = \mathop {\lim }\limits_{x \to + \infty } \frac{{2x}}{{\sqrt[3]{{{{(8{x^3} + 2x)}^2}}} + 2x\sqrt[3]{{8{x^3} + 2x}} + 4{x^2}}} = 0\)
A. + ∞
B. - ∞
C. \(\frac{4}{3}\)
D. 0
Ta có: \(H = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {16{x^4} + 3x + 1} - (4{x^2} + 2)}}{{\sqrt[4]{{16{x^4} + 3x + 1}} + \sqrt {4{x^2} + 2} }}\)
\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{16{x^4} + 3x + 1 - {{(4{x^2} + 2)}^2}}}{{\left( {\sqrt[4]{{16{x^4} + 3x + 1}} + \sqrt {4{x^2} + 2} } \right)\left( {\sqrt {16{x^4} + 3x + 1} + 4{x^2} + 2} \right)}}\)
\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 16{x^2} + 3x - 3}}{{\left( {\sqrt[4]{{16{x^4} + 3x + 1}} + \sqrt {4{x^2} + 2} } \right)\left( {\sqrt {16{x^4} + 3x + 1} + 4{x^2} + 2} \right)}}\)
Suy ra \(H = 0\).
\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{16{x^4} + 3x + 1 - {{(4{x^2} + 2)}^2}}}{{\left( {\sqrt[4]{{16{x^4} + 3x + 1}} + \sqrt {4{x^2} + 2} } \right)\left( {\sqrt {16{x^4} + 3x + 1} + 4{x^2} + 2} \right)}}\)
\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 16{x^2} + 3x - 3}}{{\left( {\sqrt[4]{{16{x^4} + 3x + 1}} + \sqrt {4{x^2} + 2} } \right)\left( {\sqrt {16{x^4} + 3x + 1} + 4{x^2} + 2} \right)}}\)
Suy ra \(H = 0\).
A. + ∞
B. - ∞
C. - 1/2
D. 0
Ta có: \(K = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 2{x^2} - x + 1 + 2\sqrt {({x^2} + 1)({x^2} - x)} }}{{\sqrt {{x^2} + 1} + \sqrt {{x^2} - x} + 2x}}\)
\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{4({x^4} - {x^3} + {x^2} - x) - {{\left( {2{x^2} + x - 1} \right)}^2}}}{{\left( {\sqrt {{x^2} + 1} + \sqrt {{x^2} - x} + 2x} \right)\left( {2\sqrt {({x^2} + 1)({x^2} - x)} + 2{x^2} + x - 1} \right)}}\)
\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{4({x^4} - {x^3} + {x^2} - x) - {{\left( {2{x^2} + x - 1} \right)}^2}}}{{\left( {\sqrt {{x^2} + 1} + \sqrt {{x^2} - x} + 2x} \right)\left( {2\sqrt {({x^2} + 1)({x^2} - x)} + 2{x^2} + x - 1} \right)}}\)
\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 8{x^3} + 7{x^2} - 2x - 1}}{{\left( {\sqrt {{x^2} + 1} + \sqrt {{x^2} - x} + 2x} \right)\left( {2\sqrt {({x^2} + 1)({x^2} - x)} + 2{x^2} + x - 1} \right)}} = - \frac{1}{2}\)
\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{4({x^4} - {x^3} + {x^2} - x) - {{\left( {2{x^2} + x - 1} \right)}^2}}}{{\left( {\sqrt {{x^2} + 1} + \sqrt {{x^2} - x} + 2x} \right)\left( {2\sqrt {({x^2} + 1)({x^2} - x)} + 2{x^2} + x - 1} \right)}}\)
\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{4({x^4} - {x^3} + {x^2} - x) - {{\left( {2{x^2} + x - 1} \right)}^2}}}{{\left( {\sqrt {{x^2} + 1} + \sqrt {{x^2} - x} + 2x} \right)\left( {2\sqrt {({x^2} + 1)({x^2} - x)} + 2{x^2} + x - 1} \right)}}\)
\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 8{x^3} + 7{x^2} - 2x - 1}}{{\left( {\sqrt {{x^2} + 1} + \sqrt {{x^2} - x} + 2x} \right)\left( {2\sqrt {({x^2} + 1)({x^2} - x)} + 2{x^2} + x - 1} \right)}} = - \frac{1}{2}\)
A. + ∞
B. - ∞
C. 3/2
D. 0
Ta có: $A = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2}(3 + \frac{5}{x} + \frac{1}{{{x^2}}})}}{{{x^2}(2 + \frac{1}{x} + \frac{1}{{{x^2}}})}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{3 + \frac{5}{x} + \frac{1}{{{x^2}}}}}{{2 + \frac{1}{x} + \frac{1}{{{x^2}}}}} = \frac{3}{2}$
A. + ∞
B. - ∞
C. \(\frac{4}{3}\)
D. Đáp án khác
Ta có: $B = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^n}({a_0} + \frac{{{a_1}}}{x} + ... + \frac{{{a_{n - 1}}}}{{{x^{n - 1}}}} + \frac{{{a_n}}}{{{x^n}}})}}{{{x^m}({b_0} + \frac{{{b_1}}}{x} + ... + \frac{{{b_{m - 1}}}}{{{x^{m - 1}}}} + \frac{{{b_m}}}{{{x^m}}})}}$
* Nếu$m = n \Rightarrow B = \mathop {\lim }\limits_{x \to + \infty } \frac{{{a_0} + \frac{{{a_1}}}{x} + ... + \frac{{{a_{n - 1}}}}{{{x^{n - 1}}}} + \frac{{{a_n}}}{{{x^n}}}}}{{{b_0} + \frac{{{b_1}}}{x} + ... + \frac{{{b_{m - 1}}}}{{{x^{m - 1}}}} + \frac{{{b_m}}}{{{x^m}}}}} = \frac{{{a_0}}}{{{b_0}}}$.
* Nếu $m > n \Rightarrow B = \mathop {\lim }\limits_{x \to + \infty } \frac{{{a_0} + \frac{{{a_1}}}{x} + ... + \frac{{{a_{n - 1}}}}{{{x^{n - 1}}}} + \frac{{{a_n}}}{{{x^n}}}}}{{{x^{m - n}}({b_0} + \frac{{{b_1}}}{x} + ... + \frac{{{b_{m - 1}}}}{{{x^{m - 1}}}} + \frac{{{b_m}}}{{{x^m}}})}} = 0$
( Vì tử$ \to {a_0}$, mẫu$ \to 0$).
* Nếu $m < n$
$ \Rightarrow B = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^{n - m}}({a_0} + \frac{{{a_1}}}{x} + ... + \frac{{{a_{n - 1}}}}{{{x^{n - 1}}}} + \frac{{{a_n}}}{{{x^n}}})}}{{{b_0} + \frac{{{b_1}}}{x} + ... + \frac{{{b_{m - 1}}}}{{{x^{m - 1}}}} + \frac{{{b_m}}}{{{x^m}}}}} = \left\{ \begin{array}{l} + \infty {\rm{ khi }}{a_0}.{b_0} > 0\\ - \infty {\rm{ khi }}{a_0}{b_0} < 0\end{array} \right.$.
* Nếu$m = n \Rightarrow B = \mathop {\lim }\limits_{x \to + \infty } \frac{{{a_0} + \frac{{{a_1}}}{x} + ... + \frac{{{a_{n - 1}}}}{{{x^{n - 1}}}} + \frac{{{a_n}}}{{{x^n}}}}}{{{b_0} + \frac{{{b_1}}}{x} + ... + \frac{{{b_{m - 1}}}}{{{x^{m - 1}}}} + \frac{{{b_m}}}{{{x^m}}}}} = \frac{{{a_0}}}{{{b_0}}}$.
* Nếu $m > n \Rightarrow B = \mathop {\lim }\limits_{x \to + \infty } \frac{{{a_0} + \frac{{{a_1}}}{x} + ... + \frac{{{a_{n - 1}}}}{{{x^{n - 1}}}} + \frac{{{a_n}}}{{{x^n}}}}}{{{x^{m - n}}({b_0} + \frac{{{b_1}}}{x} + ... + \frac{{{b_{m - 1}}}}{{{x^{m - 1}}}} + \frac{{{b_m}}}{{{x^m}}})}} = 0$
( Vì tử$ \to {a_0}$, mẫu$ \to 0$).
* Nếu $m < n$
$ \Rightarrow B = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^{n - m}}({a_0} + \frac{{{a_1}}}{x} + ... + \frac{{{a_{n - 1}}}}{{{x^{n - 1}}}} + \frac{{{a_n}}}{{{x^n}}})}}{{{b_0} + \frac{{{b_1}}}{x} + ... + \frac{{{b_{m - 1}}}}{{{x^{m - 1}}}} + \frac{{{b_m}}}{{{x^m}}}}} = \left\{ \begin{array}{l} + \infty {\rm{ khi }}{a_0}.{b_0} > 0\\ - \infty {\rm{ khi }}{a_0}{b_0} < 0\end{array} \right.$.
A. + ∞
B. - ∞
C. \( - \frac{{\sqrt[3]{3} + \sqrt 2 }}{{\sqrt 2 }}\)
D. 0
Ta có:$A = \mathop {\lim }\limits_{x \to - \infty } \frac{{x\sqrt[3]{{3 + \frac{1}{{{x^3}}}}} + x\sqrt {2 + \frac{1}{x} + \frac{1}{{{x^2}}}} }}{{ - x\sqrt[4]{{4 + \frac{2}{{{x^4}}}}}}} = - \frac{{\sqrt[3]{3} + \sqrt 2 }}{{\sqrt 2 }}$.
A. + ∞
B. - ∞
C. \(\frac{4}{3}\)
D. 0
$B = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2}(\sqrt {1 + \frac{1}{{{x^2}}}} - \frac{2}{x} + \frac{1}{{{x^2}}})}}{{x(\sqrt[3]{{2 - \frac{2}{{{x^3}}}}} + \frac{1}{x})}} = \frac{{x(\sqrt {1 + \frac{1}{{{x^2}}}} - \frac{2}{x} + \frac{1}{{{x^2}}})}}{{\sqrt[3]{{2 - \frac{2}{{{x^3}}}}} + \frac{1}{x}}} = + \infty $
(do tử \( \to + \infty \), mẫu$ \to \sqrt[3]{2}$).
(do tử \( \to + \infty \), mẫu$ \to \sqrt[3]{2}$).
A. + ∞
B. - ∞
C. \( - \frac{1}{{16}}\)
D. 0
\(A = \mathop {\lim }\limits_{x \to + \infty } \frac{{{{\left( {2 + \frac{1}{x}} \right)}^3}{{\left( {1 + \frac{2}{x}} \right)}^4}}}{{{{\left( {\frac{3}{x} - 2} \right)}^7}}} = - \frac{1}{{16}}\)
A. + ∞
B. - ∞
C. 2
D. 0
\(B = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - \sqrt {4 - \frac{3}{x} + \frac{4}{{{x^2}}}} - 2}}{{ - \sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} - x}} = 2\)
A. + ∞
B. - ∞
C. \(\frac{{2 + \sqrt 3 }}{4}\)
D. 0
\(C = \mathop {\lim }\limits_{x \to + \infty } \frac{{2 + \sqrt {3 + \frac{2}{{{x^2}}}} }}{{5 - \sqrt {1 + \frac{1}{{{x^2}}}} }} = \frac{{2 + \sqrt 3 }}{4}\)
A. + ∞
B. - ∞
C. \(\frac{4}{3}\)
D. - 1
\(D = \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt[3]{{\frac{1}{{{x^6}}} + \frac{1}{{{x^2}}} + 1}}}}{{ - \sqrt {1 + \frac{1}{x} + \frac{1}{{{x^4}}}} }} = - 1\)
A. + ∞
B. - ∞
C. \(\frac{4}{3}\)
D. 0
Ta có: \(A = \mathop {\lim }\limits_{x \to + \infty } \left( {\left| x \right|\sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} - x\sqrt[3]{{2 + \frac{1}{{{x^2}}} - \frac{1}{{{x^3}}}}}} \right)\)
\( = \mathop {\lim }\limits_{x \to + \infty } x\left( {\sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} - \sqrt[3]{{2 + \frac{1}{{{x^2}}} - \frac{1}{{{x^3}}}}}} \right) = - \infty \)
\( = \mathop {\lim }\limits_{x \to + \infty } x\left( {\sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} - \sqrt[3]{{2 + \frac{1}{{{x^2}}} - \frac{1}{{{x^3}}}}}} \right) = - \infty \)
A. + ∞
B. - ∞
C. 1/2
D. 0
Ta có: \(C = \mathop {\lim }\limits_{x \to + \infty } \frac{{x + 1}}{{\sqrt {4{x^2} + x + 1} + 2x}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{x\left( {1 + \frac{1}{x}} \right)}}{{\left| x \right|\sqrt {4 + \frac{1}{x} + \frac{1}{{{x^2}}}} + 2x}}\)\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{1 + \frac{1}{x}}}{{\sqrt {4 + \frac{1}{x} + \frac{1}{{{x^2}}}} + 2}} = \frac{1}{2}\).
A. $ + \infty $
B. - ∞
C. - 1/6
D. 0
Ta có:
\(D = \mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt[3]{{{x^3} + {x^2} + 1}} - x} \right) + \mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + x + 1} + x} \right) = M + N\)
\(M = \mathop {\lim }\limits_{x \to - \infty } \frac{{{x^2} + 1}}{{\sqrt[3]{{{{({x^3} + {x^2} + 1)}^2}}} + x.\sqrt[3]{{{x^3} + {x^2} + 1}} + {x^2}}} = \frac{1}{3}\)
\(N = \mathop {\lim }\limits_{x \to - \infty } \frac{{x + 1}}{{\sqrt {{x^2} + x + 1} - x}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{1 + \frac{1}{x}}}{{ - \sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} - 1}} = - \frac{1}{2}\)
Do đó:\(B = \frac{1}{3} - \frac{1}{2} = - \frac{1}{6}\).
\(D = \mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt[3]{{{x^3} + {x^2} + 1}} - x} \right) + \mathop {\lim }\limits_{x \to - \infty } \left( {\sqrt {{x^2} + x + 1} + x} \right) = M + N\)
\(M = \mathop {\lim }\limits_{x \to - \infty } \frac{{{x^2} + 1}}{{\sqrt[3]{{{{({x^3} + {x^2} + 1)}^2}}} + x.\sqrt[3]{{{x^3} + {x^2} + 1}} + {x^2}}} = \frac{1}{3}\)
\(N = \mathop {\lim }\limits_{x \to - \infty } \frac{{x + 1}}{{\sqrt {{x^2} + x + 1} - x}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{1 + \frac{1}{x}}}{{ - \sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} - 1}} = - \frac{1}{2}\)
Do đó:\(B = \frac{1}{3} - \frac{1}{2} = - \frac{1}{6}\).
A. + ∞
B. - ∞
C. 3/2
D. 0
Ta có: $\sqrt {{x^2} + x + 1} - 2\sqrt {{x^2} - x} + x = \frac{{{{\left( {\sqrt {{x^2} + x + 1} + x} \right)}^2} - 4({x^2} - x)}}{{\sqrt {{x^2} + x + 1} + 2\sqrt {{x^2} - x} + x}}$
\( = \frac{{2x\sqrt {{x^2} + x + 1} + 1 + 5x - 2{x^2}}}{{\sqrt {{x^2} + x + 1} + 2\sqrt {{x^2} - x} + x}}\)
\( = \frac{{2x\left( {\sqrt {{x^2} + x + 1} - x} \right)}}{{\sqrt {{x^2} + x + 1} + 2\sqrt {{x^2} - x} + x}} + \frac{{1 + 5x}}{{\sqrt {{x^2} + x + 1} + 2\sqrt {{x^2} - x} + x}}\)
\( = \frac{{2x(x + 1)}}{{\left( {\sqrt {{x^2} + x + 1} + 2\sqrt {{x^2} - x} + x} \right)\left( {\sqrt {{x^2} + x + 1} + x} \right)}} + \)
\( + \frac{{1 + 5x}}{{\sqrt {{x^2} + x + 1} + 2\sqrt {{x^2} - x} + x}}\).
Do đó: $A = \mathop {\lim }\limits_{x \to + \infty } \frac{{2 + \frac{2}{x}}}{{\left( {\sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} + 2\sqrt {1 - \frac{1}{x}} + 1} \right)\left( {\sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} + 1} \right)}} + \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{1}{x} + 5}}{{\sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} + 2\sqrt {1 - \frac{1}{x}} + 1}} = \frac{1}{4} + \frac{5}{4} = \frac{3}{2}$
\( = \frac{{2x\sqrt {{x^2} + x + 1} + 1 + 5x - 2{x^2}}}{{\sqrt {{x^2} + x + 1} + 2\sqrt {{x^2} - x} + x}}\)
\( = \frac{{2x\left( {\sqrt {{x^2} + x + 1} - x} \right)}}{{\sqrt {{x^2} + x + 1} + 2\sqrt {{x^2} - x} + x}} + \frac{{1 + 5x}}{{\sqrt {{x^2} + x + 1} + 2\sqrt {{x^2} - x} + x}}\)
\( = \frac{{2x(x + 1)}}{{\left( {\sqrt {{x^2} + x + 1} + 2\sqrt {{x^2} - x} + x} \right)\left( {\sqrt {{x^2} + x + 1} + x} \right)}} + \)
\( + \frac{{1 + 5x}}{{\sqrt {{x^2} + x + 1} + 2\sqrt {{x^2} - x} + x}}\).
Do đó: $A = \mathop {\lim }\limits_{x \to + \infty } \frac{{2 + \frac{2}{x}}}{{\left( {\sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} + 2\sqrt {1 - \frac{1}{x}} + 1} \right)\left( {\sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} + 1} \right)}} + \mathop {\lim }\limits_{x \to + \infty } \frac{{\frac{1}{x} + 5}}{{\sqrt {1 + \frac{1}{x} + \frac{1}{{{x^2}}}} + 2\sqrt {1 - \frac{1}{x}} + 1}} = \frac{1}{4} + \frac{5}{4} = \frac{3}{2}$
A. + ∞
B. - ∞
C. \( - \frac{1}{4}\)
D. 0
Ta có: \(\sqrt {{x^2} + 2x} - 2\sqrt {{x^2} + x} + x = \frac{{2{x^2} + 2x + 2x\sqrt {{x^2} + 2x} - 4{x^2} - 4x}}{{\sqrt {{x^2} + 2x} + 2\sqrt {{x^2} + x} + x}}\)
\( = 2x\frac{{\sqrt {{x^2} + 2x} - x - 1}}{{\sqrt {{x^2} + 2x} + 2\sqrt {{x^2} + x} + x}}\)
\( = \frac{{ - 2x}}{{(\sqrt {{x^2} + 2x} + 2\sqrt {{x^2} + x} + x)(\sqrt {{x^2} + 2x} + x + 1)}}\).
Nên \(B = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 2{x^2}}}{{(\sqrt {{x^2} + 2x} + 2\sqrt {{x^2} + x} + x)(\sqrt {{x^2} + 2x} + x + 1)}}\)
\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 2}}{{(\sqrt {1 + \frac{2}{x}} + 2\sqrt {1 + \frac{1}{x}} + 1)(\sqrt {1 + \frac{2}{x}} + 1 + \frac{1}{x})}} = - \frac{1}{4}\).
\( = 2x\frac{{\sqrt {{x^2} + 2x} - x - 1}}{{\sqrt {{x^2} + 2x} + 2\sqrt {{x^2} + x} + x}}\)
\( = \frac{{ - 2x}}{{(\sqrt {{x^2} + 2x} + 2\sqrt {{x^2} + x} + x)(\sqrt {{x^2} + 2x} + x + 1)}}\).
Nên \(B = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 2{x^2}}}{{(\sqrt {{x^2} + 2x} + 2\sqrt {{x^2} + x} + x)(\sqrt {{x^2} + 2x} + x + 1)}}\)
\( = \mathop {\lim }\limits_{x \to + \infty } \frac{{ - 2}}{{(\sqrt {1 + \frac{2}{x}} + 2\sqrt {1 + \frac{1}{x}} + 1)(\sqrt {1 + \frac{2}{x}} + 1 + \frac{1}{x})}} = - \frac{1}{4}\).
A. + ∞
B. - ∞
C. \(\frac{4}{3}\)
D. Đáp án khác
Ta có: \(A = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^n}({a_0} + \frac{{{a_1}}}{x} + ... + \frac{{{a_{n - 1}}}}{{{x^{n - 1}}}} + \frac{{{a_n}}}{{{x^n}}})}}{{{x^m}({b_0} + \frac{{{b_1}}}{x} + ... + \frac{{{b_{m - 1}}}}{{{x^{m - 1}}}} + \frac{{{b_m}}}{{{x^m}}})}}\)
Nếu \(m = n \Rightarrow B = \mathop {\lim }\limits_{x \to + \infty } \frac{{{a_0} + \frac{{{a_1}}}{x} + ... + \frac{{{a_{n - 1}}}}{{{x^{n - 1}}}} + \frac{{{a_n}}}{{{x^n}}}}}{{{b_0} + \frac{{{b_1}}}{x} + ... + \frac{{{b_{m - 1}}}}{{{x^{m - 1}}}} + \frac{{{b_m}}}{{{x^m}}}}} = \frac{{{a_0}}}{{{b_0}}}\).
Nếu \(m > n \Rightarrow B = \mathop {\lim }\limits_{x \to + \infty } \frac{{{a_0} + \frac{{{a_1}}}{x} + ... + \frac{{{a_{n - 1}}}}{{{x^{n - 1}}}} + \frac{{{a_n}}}{{{x^n}}}}}{{{x^{m - n}}({b_0} + \frac{{{b_1}}}{x} + ... + \frac{{{b_{m - 1}}}}{{{x^{m - 1}}}} + \frac{{{b_m}}}{{{x^m}}})}} = 0\)
( Vì tử \( \to {a_0}\), mẫu \( \to 0\)).
Nếu \(m < n\), ta có:\(B = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^{n - m}}({a_0} + \frac{{{a_1}}}{x} + ... + \frac{{{a_{n - 1}}}}{{{x^{n - 1}}}} + \frac{{{a_n}}}{{{x^n}}})}}{{{b_0} + \frac{{{b_1}}}{x} + ... + \frac{{{b_{m - 1}}}}{{{x^{m - 1}}}} + \frac{{{b_m}}}{{{x^m}}}}} = \left\{ \begin{array}{l} + \infty {\rm{ khi }}{a_0}.{b_0} > 0\\ - \infty {\rm{ khi }}{a_0}{b_0} < 0\end{array} \right.\)
Nếu \(m = n \Rightarrow B = \mathop {\lim }\limits_{x \to + \infty } \frac{{{a_0} + \frac{{{a_1}}}{x} + ... + \frac{{{a_{n - 1}}}}{{{x^{n - 1}}}} + \frac{{{a_n}}}{{{x^n}}}}}{{{b_0} + \frac{{{b_1}}}{x} + ... + \frac{{{b_{m - 1}}}}{{{x^{m - 1}}}} + \frac{{{b_m}}}{{{x^m}}}}} = \frac{{{a_0}}}{{{b_0}}}\).
Nếu \(m > n \Rightarrow B = \mathop {\lim }\limits_{x \to + \infty } \frac{{{a_0} + \frac{{{a_1}}}{x} + ... + \frac{{{a_{n - 1}}}}{{{x^{n - 1}}}} + \frac{{{a_n}}}{{{x^n}}}}}{{{x^{m - n}}({b_0} + \frac{{{b_1}}}{x} + ... + \frac{{{b_{m - 1}}}}{{{x^{m - 1}}}} + \frac{{{b_m}}}{{{x^m}}})}} = 0\)
( Vì tử \( \to {a_0}\), mẫu \( \to 0\)).
Nếu \(m < n\), ta có:\(B = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^{n - m}}({a_0} + \frac{{{a_1}}}{x} + ... + \frac{{{a_{n - 1}}}}{{{x^{n - 1}}}} + \frac{{{a_n}}}{{{x^n}}})}}{{{b_0} + \frac{{{b_1}}}{x} + ... + \frac{{{b_{m - 1}}}}{{{x^{m - 1}}}} + \frac{{{b_m}}}{{{x^m}}}}} = \left\{ \begin{array}{l} + \infty {\rm{ khi }}{a_0}.{b_0} > 0\\ - \infty {\rm{ khi }}{a_0}{b_0} < 0\end{array} \right.\)
A. + ∞
B. - ∞
C. \(\frac{4}{3}\)
D. 4
Ta có:\(B = \mathop {\lim }\limits_{x \to + \infty } \frac{{\left| x \right|\sqrt {4 + \frac{1}{x}} + x.\sqrt[3]{{8 + \frac{1}{{{x^2}}} - \frac{1}{{{x^3}}}}}}}{{\left| x \right|\sqrt[4]{{1 + \frac{3}{{{x^4}}}}}}} = \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {4 + \frac{1}{x}} + \sqrt[3]{{8 + \frac{1}{{{x^2}}} - \frac{1}{{{x^3}}}}}}}{{\sqrt[4]{{1 + \frac{3}{{{x^4}}}}}}} = 4\)
A. + ∞
B. - ∞
C. 3/2
D. 0
Ta có: \(C = \mathop {\lim }\limits_{x \to - \infty } \frac{{\left| x \right|\sqrt {4 - \frac{2}{{{x^2}}}} + \left| x \right|\sqrt[3]{{1 + \frac{1}{{{x^3}}}}}}}{{\left| x \right|\sqrt {1 + \frac{1}{{{x^2}}}} - x}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - \sqrt {4 - \frac{2}{{{x^2}}}} - \sqrt[3]{{1 + \frac{1}{{{x^3}}}}}}}{{ - \left( {\sqrt {1 + \frac{1}{{{x^2}}}} + 1} \right)}} = \frac{3}{2}\)
A. + ∞
B. - ∞
C. \(\frac{4}{3}\)
D. 0
Ta có:\(D = \mathop {\lim }\limits_{x \to + \infty } \frac{{{x^2}\left( {\sqrt {1 + \frac{1}{{{x^2}}}} + \frac{2}{x} + \frac{1}{{{x^2}}}} \right)}}{{{x^2}\left( {\sqrt[3]{{\frac{2}{{{x^3}}} + \frac{1}{{{x^5}}} + \frac{1}{{{x^6}}}}} + \frac{1}{x}} \right)}} = + \infty \).
A. Không tồn tại.
B. 0.
C. 1.
D. + ∞.
Chọn B.
Cách 1: \(0 \le \left| {\cos \frac{2}{{nx}}} \right| \le 1 \Leftrightarrow 0 \le \left| {{x^2}\cos \frac{2}{{nx}}} \right| \le {x^2}\)
Mà \(\mathop {\lim }\limits_{x \to 0} {x^2} = 0\) nên \(\mathop {\lim }\limits_{x \to 0} {x^2}\cos \frac{2}{{nx}} = 0\)
Cách 2: Bấm máy tính như sau: Chuyển qua chế độ Rad + \({x^2}\cos \frac{2}{{nx}}\) + CACL + \(x = {10^{ - 9}}\) +\(n = 10\) và so sánh
Cách 1: \(0 \le \left| {\cos \frac{2}{{nx}}} \right| \le 1 \Leftrightarrow 0 \le \left| {{x^2}\cos \frac{2}{{nx}}} \right| \le {x^2}\)
Mà \(\mathop {\lim }\limits_{x \to 0} {x^2} = 0\) nên \(\mathop {\lim }\limits_{x \to 0} {x^2}\cos \frac{2}{{nx}} = 0\)
Cách 2: Bấm máy tính như sau: Chuyển qua chế độ Rad + \({x^2}\cos \frac{2}{{nx}}\) + CACL + \(x = {10^{ - 9}}\) +\(n = 10\) và so sánh
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