Đáp án:
Giải thích các bước giải:
`M=(\sqrt{x}-\frac{x+2}{\sqrt{x}+1}):(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}-4}{x-1})`
`M=[\frac{\sqrt{x}(\sqrt{x}+1)}{\sqrt{x}+1}-\frac{x+2}{\sqrt{x}+1}]:[\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}+\frac{\sqrt{x}-4}{(\sqrt{x}-1)(\sqrt{x}+1)}]`
`M=\frac{x+\sqrt{x}-x-2}{\sqrt{x}+1}.\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{x-\sqrt{x}+\sqrt{x}-4)`
`M=\frac{\sqrt{x}-2}{\sqrt{x}+1}.\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{(\sqrt{x}-2)(\sqrt{x}+2)}`
`M=\frac{\sqrt{x}-1}{\sqrt{x}+2}`
`M=\frac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\frac{3}{\sqrt{x}+2}`
Ta có: `\sqrt{x}+2 \ge 2 \forall x`
`⇔ \frac{1}{\sqrt{x}+2} \le \frac{1}{2}`
`⇔ \frac{3}{\sqrt{x}+2} \le \frac{3}{2}`
`⇒ C \ge 1-3/2=-1/2`
Dấu `=` xảy ra khi `\sqrt{x}+2 =2⇔x=0\ (TM)`
Vậy `C_{min}=-1/2` khi `x=0`
