Đáp án:
- a) $M=\dfrac{3\sqrt{x}}{\sqrt{x}-3}$
- b) $0\,<\,x\,<\,36$
- c) $0\,<\,x\,<\,9$
- 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)
$\,\,\,\,\,\,\,M=\dfrac{-3}{2\sqrt{x}}$
$\Leftrightarrow \dfrac{3\sqrt{x}}{\sqrt{x}-3}=\dfrac{-3}{2\sqrt{x}}$
$\Leftrightarrow 3\sqrt{x}\,.\,2\sqrt{x}=-3\left( \sqrt{x}-3 \right)$
$\Leftrightarrow 6x=-3\sqrt{x}+9$
$\Leftrightarrow 6x+3\sqrt{x}-9=0$
$\Leftrightarrow 2x+\sqrt{x}-3=0$
$\Leftrightarrow \left( \sqrt{x}-1 \right)\left( 2\sqrt{x}+3 \right)=0$
$\Leftrightarrow\left[\begin{array}{l}\sqrt{x}=1\\\sqrt{x}=-\dfrac{3}{2}\,\,\,\left(\text{ vô lý }\right)\end{array}\right.$
$\Leftrightarrow x=1\,\,\,\left(\text{ nhận }\right)$
