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giải đáp
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Giúp với!!!!!!!!!!
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$= \mathop {\lim \limits_{x \to 0} \frac{sin(\frac{\pi}2-\frac{\pi}2.cosx)}{sin^2(\frac{x}2)}}$ $= \mathop {\lim \limits_{x \to 0} \frac{sin(\frac{\pi}2(1-cosx))}{sin^2(\frac{x}2)}}$ $= \mathop {\lim \limits_{x \to 0} \frac{sin(\frac{\pi}2(2.sin^2(\frac{\pi}2))}{sin^2(\frac{x}2)}}$ $=\mathop {\lim }\limits_{x \to 0} \frac{sin(\pi.sin^2(\frac{x}2))}{sin^2(\frac{x}2)}$ $=\mathop {\lim }\limits_{x \to 0} \frac{sin(\pi.sin^2(\frac{x}2))}{\pi.sin^2(\frac{x}2)}.\pi=\pi$
(vì $\mathop {\lim }\limits_{x \to 0} \frac{sinax}{ax}=1$) |
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giải đáp
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Giúp với!!!!!!!!
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$4cosx.cos2x.cos3x=2.cos3x(cos3x+cosx)=2cos^23x.2cosx.cos3x$ $=1+cos6x+cos4x+cos2x$
đề bài $=\mathop {\lim \limits_{x \to 0} \frac{4-1-cos6x-cos4x-cos2x}{1-cosx}}$
$=\mathop {\lim \limits_{x \to 0} \frac{1-cos6x+1-cos4x+1-cos2x}{1-cosx}}$
$=\mathop {\lim \limits_{x \to 0} \frac{\frac{1-cos6x+1-cos4x+1-cos2x}{x^2}}{\frac{1-cosx}{x^2}}} (1)$
Ta có $\mathop {\lim }\limits_{x \to 0} \frac{1-cosax}{x^2}=\frac{a^2}{2}$ (tự cm đc nha bạn, dễ lắm ) Thế vào ta có $(1)= \frac{\frac{6^2}2 +\frac{4^2}2+\frac{2^2}2}{\frac{1}2}=56$ |
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giải đáp
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Giúp với!!!!!
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Xét $cos(x+\frac{\pi}6)=\frac{\sqrt 3 cosx}2-\frac{sinx}2=\frac{\sqrt 3cosx -sinx}2$ $tan^3x-3tanx=tanx(tan^2x-3)=tanx(tanx-\sqrt 3)(tanx+\sqrt 3)=tanx(\frac{sinx-\sqrt 3cosx}{cosx})(\frac{sinx + \sqrt 3cosx}{cosx})$
$Đề bài=\mathop {\lim \limits_{x \to \frac{\pi}3} \frac{2tanx(sinx-\sqrt 3cosx)(sinx+\sqrt 3cosx)}{cos^2x(\sqrt 3cosx-sinx)} }$
$=\mathop {\lim \limits_{x \to \frac{\pi}3} \frac{(-2tanx)(sinx+\sqrt 3cosx)}{cos^2x}} = -24$
Nếu thấy đúng bạn nhấn V và vote up hộ mình. Cảm ơn :) |
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giải đáp
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Giúp với!!!!!
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$4.cos3x.cos5x.cos7x=2.cos5x.(cos10x+cos4x)$$=cos15+cos5x+cos9x+cosx $
Tiếp theo sử dụng định lí: $\mathop {\lim \limits_{x \to 0} \frac{sinax}a=1}$
ta tính giới hạn tổng quát sau: $ \mathop {\lim \limits_{x \to 0} \frac{1-cosax}{x^2}=\frac{a^2}2}$ (tự cm nha bạn, ko khó đâu )
Vậy giới hạn cần tìm bằng: limx→09883.4−cos15x−cos9x−cos5x−cosx4.sin27x=limx→09883.1−cos15xx2+1−cos9xx2+1−cos5xx2+1−cosxx2sin27xx2=9883.1522+922+522+1272=9883.8398=1 $\mathop {\lim \limits_{x \to 0} }\frac{98}{83}.\frac{4-cos15x-cos9x-cos5x-cosx}{4.sin^27x}$
$=\mathop {\lim\limits_{x \to 0}} \frac{98}{83}.\frac{\frac{1-cos15x+1-cos9x+1-cos5x+1-cosx}{x^2}}{4.\frac{sin^27x}{x^2}}$
$=\frac{98}{83}.\frac{\frac{15^2}2+\frac{9^2}2+\frac{5^2}2+\frac{1}2}{4.7^2}=1$
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giải đáp
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giới hạn 11 khó tt
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c) $= \mathop {\lim }\limits_{x \to 2} \frac{[(x-2)(x+1)]^{20}}{[(x-4)(x-2)^2]^{10}}$ $= \mathop {\lim }\limits_{x \to 2} \frac{(x-2)^{20}.(x+1)^{20}}{(x-4)^{10}.(x-2)^{20}}$
$= \mathop {\lim }\limits_{x \to 2} \frac{(x+1)^{20}}{(x-4)^{10}}=\frac{3^{20}}{2^{10}}$ |
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giải đáp
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giới hạn 11 khó tt
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b. $= \mathop {\lim }\limits_{x \to 1} \frac{(x^{99}+x^{98}+...+x-1)(x-1)}{(x^{49}+x^{48}+...+x-1)(x-1)}$ $=\mathop {\lim }\limits_{x \to 1} \frac{x^{100}+x^{99}+...+x-1}{x^{49}+x^{48}+...+x-1}$
$= \frac{1^{99}+1^{98}+...+1-1}{1^{49}+1^{48}+...+1-1}=\frac{98}{48}=\frac{49}{24}$ |
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giải đáp
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giới hạn 11 khó tt
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$a) = \mathop {\lim }\limits_{x \to 0} [\frac{\sqrt{x+1}-1}{x}-\frac{\sqrt[3]{x+1}-1}x]$ $=\mathop {\lim }\limits_{x \to 0} [\frac{x}{x(\sqrt{x+1}+1)}-\frac{x}{x(\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}+1)}]$
$= \mathop {\lim }\limits_{x \to 0} [\frac{1}{\sqrt{x+1}+1}-\frac{1}{\sqrt[3]{(x+1)^2}+\sqrt[3]{x+1}+1}]$
$=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$ |
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giải đáp
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tính giới hạn
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$= \mathop {\lim }\limits_{x \to +\infty } \sqrt{\frac{x^3(x^2+2x+1)}{2x^4+x^2+1}}$ $= \mathop {\lim }\limits_{x \to +\infty } \sqrt{\frac{x^5+2x^4+x^3}{2x^4+x^2+1}}$ $= \mathop {\lim }\limits_{x \to +\infty } \sqrt{\frac{x(1+\frac{2}{x}+\frac{1}{x^2})}{2+\frac{1}{x^2}+\frac{1}{x^4}}}$ $= \sqrt{+\infty . \frac{1}{2}}= +\infty$ |
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giải đáp
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tính giới hạn
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$= \mathop {\lim }\limits_{x \to +\infty } \sqrt{\frac{(x-1)(x^2+4x+4)}{x^3-1}}$ $= \mathop {\lim }\limits_{x \to +\infty } \sqrt{\frac{x^3+3x^2-4}{x^3-1}}$
$= \mathop {\lim }\limits_{x \to +\infty } \sqrt{\frac{1+\frac{3}{x}-\frac{4}{x^3}}{1-\frac{1}{x^3}}}$
$= \sqrt{\frac{1}{1}}=1$ |
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giải đáp
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tính giới hạn
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$= \mathop {\lim }\limits_{x \to +\infty } \sqrt{\frac{x.(x+1)^2}{2x^4+x^2+1}}$ $= \mathop {\lim }\limits_{x \to +\infty } \sqrt{\frac{x^3+2x^2+x}{2x^4+x^2+1}}$
$= \mathop {\lim }\limits_{x \to +\infty } \sqrt{\frac{\frac{1}{x}+\frac{2}{x^2}+\frac{1}{x^3}}{2+\frac{1}{x^2}+\frac{1}{x^4}}}$
$= \sqrt{\frac{0}{2}}=0$ |
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giải đáp
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tính giới hạn
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$=\mathop {\lim }\limits_{x \to +\infty } \sqrt[3] {\frac{2+\frac{1}{x^2}-\frac{1}{x^5}}{(2-\frac{1}{x^2})(1+\frac{1}{x^2})}}$ $= \sqrt[3]{\frac{2+0+0}{(2-0)(1+0)}}=1$ |
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giải đáp
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Giúp em với, toán 9 nè.
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Xét $y(x+2)=13+3x$
$\Rightarrow y=\frac{13+3x}{x+2}=\frac{3x+6+7}{x+2}=3+\frac{7}{x+2}$
Để $y \in Z \Rightarrow \frac{7}{x+2} \in Z \Rightarrow x+2 \in { \pm 1; \pm 7}$
$\Rightarrow x \in {-9;-3;-1;5}$
Vậy $(x;y)= (-9;2); (-3;-4); (-1;10); (5;4)$
Nếu thấy bài đúng bạn nhấn V và vote úp nha. Cảm ơn bạn :)
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giải đáp
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Tìm lim biết
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c) Xét $lim (\sqrt{4n^2+3n+1}-2n) = lim \frac{3n+1}{\sqrt{4n^2+3n+1}+2n} = lim \frac{3+\frac{1}n}{\sqrt{4+\frac{3}n+\frac{1}{n^2}}+2} = \frac{3}{4}$
$lim (\sqrt[3]{8n^3+2n-1} -2n)= lim \frac{2n-1}{\sqrt[3]{(8n^3+2n-1)^2}+2n.\sqrt[3]{8n^3+2n-1}+4n^2}= lim \frac{\frac{2}n - \frac{1}{n^2}}{\sqrt[3]{(8+\frac{2}{n^2}-\frac{1}{n^3})^2}+2.\sqrt[3]{8+\frac{2}{n^2}-\frac{1}{n^3}}+4}=\frac{0}{\sqrt[3]{64}+2.\sqrt[3]{8}+4}=0$
vì $n \to + \infty \Rightarrow \frac{2}{n}-\frac{1}{n^2} >0 $
$\Rightarrow lim \frac{\frac{2}n - \frac{1}{n^2}}{\sqrt[3]{(8+\frac{2}{n^2}-\frac{1}{n^3})^2}+2.\sqrt[3]{8+\frac{2}{n^2}-\frac{1}{n^3}}+4} = 0^+$
$\Rightarrow lim \sqrt[3]{8n^3+2n-1}-2n = 0^+$
$lim u_n=\frac{\frac{3}4}{0^+} = + \infty$ |
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giải đáp
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Tìm lim biết
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b) $lim u_n=\frac{1+2.\sqrt{\frac {1}{n^3}}+\frac {3}{n^2}}{2+\frac{1}n -\sqrt{\frac 1{n^3}}}=\frac 1{2}$
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giải đáp
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ne` la de ktra 1 tiet, mn co' hung' thu' thi` vao` lam` nha:)
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Câu 3 Đặt $\left\{ \begin{array}{l} V_{17}=a\\ V_{20}=b \end{array} \right.$
$\Rightarrow \left\{ \begin{array}{l} a-b=9\\ a^2+b^2=153 \end{array} \right.$
$\Rightarrow \left\{ \begin{array}{l} ab=36\\ a+b=15 \end{array} \right.$
$\Rightarrow \left\{ \begin{array}{l} a=12\\ b=3 \end{array} \right.$
$\Rightarrow \left\{ \begin{array}{l} V_0.q^{16}=12\\ V_0.q^{19}=3 \end{array} \right.$
Tính ra $V_0, q$ là ra $S_{11}$ |
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