Giải giúp mk 2 câu này vs ạ , c ơn

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Giải giúp mk 2 câu này vs ạ , c ơn
giai-giup-mk-2-cau-nay-vs-a-c-on

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King 5 years 2021-05-20T07:10:22+00:00 1 Answers 22 views 0

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  1. kiet
    0
    2021-05-20T07:12:02+00:00
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    Đáp án:

    79) $\lim\limits_{x\to -2}\dfrac{x^4-16}{x^3 + 2x^2}=-8$

    80) $\lim\limits_{x\to 0}\dfrac{\sqrt{5+x} – \sqrt{5-x}}{x}=\dfrac{\sqrt5}{5}$

    Giải thích các bước giải:

    79) $\lim\limits_{x\to -2}\dfrac{x^4-16}{x^3 + 2x^2}$

    $= \lim\limits_{x\to -2}\dfrac{(x+2)(x-2)(x^2 +4)}{x^2(x+2)}$

    $=\lim\limits_{x\to -2}\dfrac{(x-2)(x^2 +4)}{x^2}$

    $=\dfrac{(-2-2)[(-2)^2 +4)}{(-2)^2}$

    $= -8$

    80) $\lim\limits_{x\to 0}\dfrac{\sqrt{5+x} – \sqrt{5-x}}{x}$

    $= \lim\limits_{x\to 0}\dfrac{\left(\sqrt{5+x} – \sqrt{5-x}\right)\left(\sqrt{5+x} +\sqrt{5-x}\right)}{x\left(\sqrt{5+x} +\sqrt{5-x}\right)}$

    $=\lim\limits_{x\to 0}\dfrac{5 + x – (5-x)}{x\left(\sqrt{5+x} +\sqrt{5-x}\right)}$

    $=\lim\limits_{x\to 0}\dfrac{2x}{x\left(\sqrt{5+x} +\sqrt{5-x}\right)}$

    $=\lim\limits_{x\to 0}\dfrac{2}{\sqrt{5+x} +\sqrt{5-x}}$

    $= \dfrac{2}{\sqrt{5+0} +\sqrt{5-0}}$

    $=\dfrac{\sqrt5}{5}$

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