Giúp em câu này nhanh với mọi ng ơi :((

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Giúp em câu này nhanh với mọi ng ơi :((
giup-em-cau-nay-nhanh-voi-moi-ng-oi

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Thiên Hương 5 years 2021-05-17T09:52:55+00:00 2 Answers 28 views 0

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  1. thuhuong
    0
    2021-05-17T09:54:00+00:00
    Reply

    Đáp án:

     

    Giải thích các bước giải:

     

    giup-em-cau-nay-nhanh-voi-moi-ng-oi

  2. minhkhang
    0
    2021-05-17T09:54:11+00:00
    Reply

    Đáp án:

    1. a) $M=\dfrac{3\sqrt{x}}{\sqrt{x}-3}$
    2. b) $0\,<\,x\,<\,36$
    3. c) $0\,<\,x\,<\,9$
    4. d) $x=1$

     

    Giải thích:

    $M=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\,\,\,\left( x>0\,,\,x\ne 9 \right)$

     

    $\bullet \,\,\,\,\,$Với $x>0\,,\,x\ne 9$, ta có:

    a)

    $M=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}$

    $M=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{\left( \sqrt{x}+3 \right)\left( \sqrt{x}-3 \right)}$

    $M=\dfrac{2\sqrt{x}\left( \sqrt{x}-3 \right)+\left( \sqrt{x}+1 \right)\left( \sqrt{x}+3 \right)+11\sqrt{x}-3}{\left( \sqrt{x}-3 \right)\left( \sqrt{x}+3 \right)}$

    $M=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}+\sqrt{x}+3+11\sqrt{x}-3}{\left( \sqrt{x}-3 \right)\left( \sqrt{x}+3 \right)}$

    $M=\dfrac{3x+9\sqrt{x}}{\left( \sqrt{x}-3 \right)\left( \sqrt{x}+3 \right)}$

    $M=\dfrac{3\sqrt{x}\left( \sqrt{x}+3 \right)}{\left( \sqrt{x}-3 \right)\left( \sqrt{x}+3 \right)}$

    $M=\dfrac{3\sqrt{x}}{\sqrt{x}-3}$

     

    b)

    $\,\,\,\,\,\,\,\dfrac{1}{M}\,<\,\dfrac{1}{6}$

    $\Leftrightarrow \dfrac{\sqrt{x}-3}{3\sqrt{x}}\,<\dfrac{1}{6}$

    $\Leftrightarrow \dfrac{\sqrt{x}-3}{3\sqrt{x}}\,-\,\dfrac{1}{6}\,<\,0$

    $\Leftrightarrow \dfrac{2\left( \sqrt{x}-3 \right)-\sqrt{x}}{6\sqrt{x}}\,<\,0$

    $\Leftrightarrow \dfrac{\sqrt{x}-6}{6\sqrt{x}}\,<\,0$

    $\Leftrightarrow \sqrt{x}-6\,<\,0$ ( vì $6\sqrt{x}\,>0$ với mọi $x>0\,,\,x\ne 9$ )

    $\Leftrightarrow \sqrt{x}\,<\,6$

    $\Leftrightarrow 0\,<\,x\,<\,36$

     

    c)

    $\,\,\,\,\,\,\,M\,<\,0$

    $\Leftrightarrow \dfrac{3\sqrt{x}}{\sqrt{x}-3}\,<\,0$

    $\Leftrightarrow \sqrt{x}-3\,<\,0$ ( vì $3\sqrt{x}\,>\,0$ với mọi $x>0\,,\,x\ne 9$ )

    $\Leftrightarrow \sqrt{x}<3$

    $\Leftrightarrow 0\,<\,x\,<\,9$

     

    d)

    $\,\,\,\,\,\,\,M=\dfrac{-3}{2\sqrt{x}}$

    $\Leftrightarrow \dfrac{3\sqrt{x}}{\sqrt{x}-3}=\dfrac{-3}{2\sqrt{x}}$

    $\Leftrightarrow 3\sqrt{x}\,.\,2\sqrt{x}=-3\left( \sqrt{x}-3 \right)$

    $\Leftrightarrow 6x=-3\sqrt{x}+9$

    $\Leftrightarrow 6x+3\sqrt{x}-9=0$

    $\Leftrightarrow 2x+\sqrt{x}-3=0$

    $\Leftrightarrow \left( \sqrt{x}-1 \right)\left( 2\sqrt{x}+3 \right)=0$

    $\Leftrightarrow\left[\begin{array}{l}\sqrt{x}=1\\\sqrt{x}=-\dfrac{3}{2}\,\,\,\left(\text{ vô lý }\right)\end{array}\right.$

    $\Leftrightarrow x=1\,\,\,\left(\text{ nhận }\right)$

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