Question

Cho P =(căn x)/(căn x-1)-(2 căn x)/(căn x+1)+(x-3)/(x-1) a) tìm ĐKXĐ b) rút gọn P

Answer 1

`a)->` ĐKXĐ : `x>=0;x\ne1`

`b)` Ta có :

`P=(\sqrtx)/(\sqrtx-1)-(2\sqrtx)/(\sqrtx+1)+(x-3)/(x-1)`

`P=(\sqrtx(\sqrtx+1)-2\sqrtx(\sqrtx-1)+x-3)/(x-1)`

`P=(x+\sqrtx-2x+2\sqrtx+x-3)/(x-1)`

`P=(3\sqrtx-3)/(x-1)`

`P=(3(\sqrtx-1))/((\sqrtx-1)(\sqrtx+1))`

`P=3/(\sqrtx+1)`

Vậy `P=3/(\sqrtx+1)` khi `x>=0;x\ne1`

Answer 2

\(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\sqrt{x}+1}+\dfrac{x-3}{x-1}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{\sqrt{x}+1}+\dfrac{x-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{x-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{x+\sqrt{x}-2x+2\sqrt{x}+x-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\\)

\(=\dfrac{3}{\sqrt{x}+1}\)

Bổ sung \(\text{đ}k\text{x}\text{đ}:x\ge0;x\ne1\)