Đáp án+Giải thích các bước giải:
`B=(\frac{\sqrt{x}}{2+\sqrt{x}}+\frac{x+4}{4-x}):(\frac{2\sqrt{x}+1}{x-2\sqrt{x}}-1/(\sqrt{x}))(x>0,x\ne4)`
`=(\frac{\sqrt{x}(2-\sqrt{x})+x+4}{(2+\sqrt{x})(2-\sqrt{x})}):(\frac{2\sqrt{x}+1}{\sqrt{x}(\sqrt{x}-2)}-\frac{1}{\sqrt{x}})`
`=(\frac{2\sqrt{x}-x+x+4}{(2+\sqrt{x})(2-\sqrt{x})}):(\frac{2\sqrt{x}+1-\sqrt{x}+2}{\sqrt{x}(\sqrt{x}-2)})`
`=\frac{2\sqrt{x}+4}{(2+\sqrt{x})(2-\sqrt{x})}.\frac{\sqrt{x}(\sqrt{x}-2)}{\sqrt{x}+3}`
`=\frac{-2(\sqrt{x}+2)}{2+\sqrt{x}}.\frac{\sqrt{x}}{\sqrt{x}+3}`
`=\frac{-2\sqrt{x}}{\sqrt{x}+3}`
Khi `B=-1`
`=>\frac{-2\sqrt{x}}{\sqrt{x}+3}=-1`
`<=>-2\sqrt{x}=-\sqrt{x}-3`
`<=>-2\sqrt{x}+\sqrt{x}+3=0`
`<=>-\sqrt{x}=-3`
`<=>\sqrt{x}=3`
`<=>x=9(tm)`
Vậy `x=9<=>B=-1`
