Cho biểu thức `A=“(\frac{\sqrt{x}}{2+\sqrt{x}}“+“\frac{x+4}{4-x})“:“(\frac{2\sqrt{x}+1}{x-2\sqrt{x}}“-“\frac{1}{\sqrt{x}})“(x>0;x`$\neq$`4)`
a) Rút gọn `A`
b) Tìm `x` để `A=-1/3`
Cho biểu thức `A=“(\frac{\sqrt{x}}{2+\sqrt{x}}“+“\frac{x+4}{4-x})“:“(\frac{2\sqrt{x}+1}{x-2\sqrt{x}}“-“\frac{1}{\sqrt{x}})“(x>0;x`$\neq$`4)`
a) Rút gọn `A`
b) Tìm `x` để `A=-1/3`
`***` Lời giải chi tiết `***`
`a)` `ĐKXĐ:x>0;x\ne 4`
`A=((\sqrt{x})/(2+\sqrt{x})+(x+4)/(4-x)):((2\sqrt{x}+1)/(x-2\sqrt{x})-(1)/(\sqrt{x}))`
`=(\sqrt{x}(\sqrt{x}-2)-(x+4))/((\sqrt{x}-2)(\sqrt{x}+2)):(2\sqrt{x}+1-(\sqrt{x}-2))/(\sqrt{x}(\sqrt{x}-2))`
`=(x-2\sqrt{x}-x-4)/((\sqrt{x}-2)(\sqrt{x}+2)).(\sqrt{x}(\sqrt{x}-2))/(2\sqrt{x}+1-\sqrt{x}+2)`
`=(-2\sqrt{x}-4)/(\sqrt{x}+2).(\sqrt{x})/(\sqrt{x}+3)`
`=(-2(\sqrt{x}+2))/(\sqrt{x}+2).(\sqrt{x})/(\sqrt{x}+3)`
`=-(2\sqrt{x})/(\sqrt{x}+3)`
`b)`
`A=-(1)/(3)`
`<=>-(2\sqrt{x})/(\sqrt{x}+3)=-(1)/(3)`
`<=>(2\sqrt{x})/(\sqrt{x}+3)=(1)/(3)`
`<=>6\sqrt{x}=\sqrt{x}+3`
`<=>5\sqrt{x}=3`
`<=>\sqrt{x}=(3)/(5)`
`<=>x=(9)/(25)\ ™`
a) Với `x>0 ; x ne 4` ta có :
`A= (\frac{\sqrt{x}}{2+\sqrt{x}}+\frac{x+4}{4-x}):(\frac{2\sqrt{x}+1}{x-2\sqrt{x}}-\frac{1}{\sqrt{x}})`
`= (\frac{\sqrt{x}(2-\sqrt{x})}{4-x}+\frac{x+4}{4-x}):(\frac{2\sqrt{x}+1}{\sqrt{x}(\sqrt{x}-2)}-\frac{\sqrt{x}-2}{\sqrt{x}(\sqrt{x}-2)})`
`= (\frac{2\sqrt{x}-x+x+4}{4-x}):(\frac{2\sqrt{x}+1-\sqrt{x}+2}{\sqrt{x}(\sqrt{x}-2)})`
`= -\frac{2(\sqrt{x}+2)}{x-4}:\frac{\sqrt{x}+3}{\sqrt{x}(\sqrt{x}-2)}`
`= -\frac{2}{\sqrt{x}-2} · \frac{\sqrt{x}(\sqrt{x}-2)}{\sqrt{x}+3}`
`= -\frac{2\sqrt{x}}{\sqrt{x}+3} `
Vậy với `x>0 ; x ne 4` thì `A=-\frac{2\sqrt{x}}{\sqrt{x}+3} `
b) Với `x>0 ; x ne 4` để `A=-1/3` thì :
`-\frac{2\sqrt{x}}{\sqrt{x}+3} = -1/3`
`<=>` `\frac{2\sqrt{x}}{\sqrt{x}+3} = 1/3`
`<=>` `\frac{6\sqrt{x}}{3(\sqrt{x}+3)} = (\sqrt{x}+3)/(3(\sqrt{x}+3))`
`=>` `6\sqrt{x} = \sqrt{x}+3`
`<=>` `5\sqrt{x} = 3`
`<=>` `\sqrt{x} = 3/5`
`<=>` `x = 9/25 ™`
Vậy `A=-1/3` khi `x = 9/25`