Cho P=$\frac{2x+2}{ √x}$ + $\frac{x √x-1}{x- √x}$ - $\frac{x √x+1}{x-+√x}$ (x>0,xkhác 1) Rút gọn P
2 câu trả lời
Đáp án+Giải thích các bước giải:
`P=\frac{2x+2}{\sqrt{x}}+\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}`
ĐKXĐ: `x>0;xne1`
`=\frac{2x+2}{\sqrt{x}}+\frac{(\sqrt{x})^3-1}{\sqrt{x}(\sqrt{x}-1)}-\frac{(\sqrt{x})^3+1}{\sqrt{x}(\sqrt{x}+1)}`
`=\frac{2x+2}{\sqrt{x}}+\frac{(\sqrt{x}-1)(x+\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}-\frac{(\sqrt{x}+1)(x-\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)}`
`=\frac{2x+2}{\sqrt{x}}+\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}`
`=\frac{2x+2+x+\sqrt{x}+1-(x-\sqrt{x}+1)}{\sqrt{x}}`
`=\frac{2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}`
`=\frac{2x+2+2\sqrt{x}}{\sqrt{x}}`
Vậy `P=\frac{2x+2+2\sqrt{x}}{\sqrt{x}}` với `x>0;xne1`
`P=(2x+2)/(\sqrt{x})+(x\sqrt{x}-1)/(x-\sqrt{x})-(x\sqrt{x}+1)/(x+\sqrt{x})` `(x>0,xne1)`
`P=(2x+2)/(\sqrt{x})+((\sqrt{x}-1)(x+\sqrt{x}+1))/(\sqrt{x}(\sqrt{x}-1))-((\sqrt{x}+1)(x-\sqrt{x}+1))/(\sqrt{x}(\sqrt{x}+1))`
`P=(2x+2)/(\sqrt{x})+(x+\sqrt{x}+1)/(\sqrt{x})-(x-\sqrt{x}+1)/(\sqrt{x})`
`P=(2x+2+x+\sqrt{x}+1-x+\sqrt{x}-1)/(\sqrt{x})`
`P=(2x+2+2\sqrt{x})/(\sqrt{x})`