Đáp án:
Giải thích các bước giải:
`M=\frac{2+\sqrt{x}}{\sqrt{x}}:(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{x+\sqrt{x}})`
ĐK: `x >0`
`M=\frac{2+\sqrt{x}}{\sqrt{x}}:(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{x+\sqrt{x}})`
`M=\frac{2+\sqrt{x}}{\sqrt{x}}:[\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)}+\frac{2\sqrt{x}+1}{\sqrt{x}(\sqrt{x}+1)}]`
`M=\frac{2+\sqrt{x}}{\sqrt{x}}:[\frac{x-1+2\sqrt{x}+1}{\sqrt{x}(\sqrt{x}+1)}]`
`M=\frac{2+\sqrt{x}}{\sqrt{x}}.\frac{\sqrt{x}(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+2)}`
`M=\frac{\sqrt{x}+1}{\sqrt{x}}`
`M>3/2`
`⇔ \frac{\sqrt{x}+1}{\sqrt{x}}>3/2`
`⇔ \frac{\sqrt{x}+1}{\sqrt{x}}-3/2>0`
`⇔ \frac{2\sqrt{x}+2-3\sqrt{x}}{2\sqrt{x}}>0`
`⇔ \frac{-\sqrt{x}+2}{2\sqrt{x}}>0`
Do `x >0 ⇒ \sqrt{x}>0⇒2\sqrt{x}>0`
`⇒ -\sqrt{x}+2>0`
`⇔ \sqrt{x}>2`
`⇔ x>4` kết hợp ĐKXĐ
Vậy `x>4` thì `M>3/2`