Question

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

chứng minh B=\(\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)

 

Answer 1

\(B=\dfrac{2\sqrt{x}-3}{\sqrt{x}-1}+\dfrac{3-\sqrt{x}}{x-1}\left(dkxd:x\ne1,x\ge0\right)\)

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

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

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

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

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

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

Answer 2

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

Answer 3

\(B=\dfrac{2\sqrt{x}-3}{\sqrt{x}-1}+\dfrac{3-\sqrt{x}}{x-1}\) ĐK: \(x\ge0;x\ne1\)

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

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

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

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

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\) (Đpcm).

Answer 4

Điều kiện: `x>=0;x\ne1`

Với `x>=0;x\ne1` ta có:
`B=(2sqrtx-3)/(sqrtx-1)+(3-sqrtx)/(x-1)`

`=((2sqrtx-3)(sqrtx+1)+(3-sqrtx))/((sqrtx-1)(sqrtx+1))`

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

`=(2x-2sqrtx)/((sqrtx-1)(sqrtx+1))`

`=(2sqrtx(sqrtx-1))/((sqrtx-1)(sqrtx+1))`

`=(2sqrtx)/(sqrtx+1)(Q.E.D)`