Đáp án:
$1)\dfrac{5}{3}\\ 2)\dfrac{2\sqrt{x}-9}{\sqrt{x}+2}\\ 3)x =9$
Giải thích các bước giải:
$1)x=16, A=\dfrac{3\sqrt{16}-2}{\sqrt{16}+2}\\ =\dfrac{3.4-2}{4+2}\\ =\dfrac{5}{3}\\ 2)B=\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{2}{\sqrt{x}}-\dfrac{7\sqrt{x}-4}{2\sqrt{x}+x}\\ =\dfrac{2\sqrt{x}.\sqrt{x}}{\sqrt{x}+2}-\dfrac{2(\sqrt{x}+2)}{\sqrt{x}}-\dfrac{7\sqrt{x}-4}{\sqrt{x}(\sqrt{x}+2)}\\ =\dfrac{2x-2(\sqrt{x}+2)-(7\sqrt{x}-4)}{\sqrt{x}(\sqrt{x}+2)}\\ =\dfrac{2x-9\sqrt{x}-8}{\sqrt{x}(\sqrt{x}+2)}\\=\dfrac{2x-9\sqrt{x}}{\sqrt{x}(\sqrt{x}+2)}\\=\dfrac{2\sqrt{x}-9}{\sqrt{x}+2}\\ 3)M=A-B\\ =\dfrac{3\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{2\sqrt{x}-9}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}+7}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}+2+5}{\sqrt{x}+2}\\ =1+\dfrac{5}{\sqrt{x}+2}\\ M \in \mathbb{Z}\Rightarrow 5 \vdots (\sqrt{x}+2)\\ \Rightarrow \sqrt{x}+2 \in \{\pm 1; \pm 5\}\\ \Rightarrow x =9$
