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\(Đkxđ\Leftrightarrow\hept{\begin{cases}x>0\\\left(\sqrt{x}-1\right)^2>0\end{cases}\Rightarrow\hept{\begin{cases}x>0\\x>1\end{cases}\Rightarrow}x>1}\)
\(C=\)\(\frac{1}{\sqrt{x}}+\frac{3}{x\sqrt{x}}+1+\frac{2}{x-\sqrt{x}+1}\)
\(=\frac{1}{\sqrt{x}}+\frac{3}{x\sqrt{x}}+1+\frac{2}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{x\left(\sqrt{x}-1\right)^2}{x\sqrt{x}\left(\sqrt{x}-1\right)^2}+\frac{3\left(\sqrt{x}-1\right)^2}{x\sqrt{x}\left(\sqrt{x}-1\right)^2}+\frac{x\sqrt{x}\left(\sqrt{x}-1\right)^2}{x\sqrt{x}\left(\sqrt{x}-1\right)^2}+\frac{2x.\sqrt{x}}{x\sqrt{x}\left(\sqrt{x-1}\right)^2}\)
\(=x\left(\sqrt{x}-1\right)^2+3\left(\sqrt{x}-1\right)^2+x\sqrt{x}\left(\sqrt{x}-1\right)^2+2x.\sqrt{x}\)
.....
`A=(1/(x-sqrtx)+1/(sqrtx-1)):(sqrtx+1)/(sqrtx-1)^2`
`=((sqrtx+1)/(x-sqrtx)).(sqrtx-1)^2/(sqrtx+1)`
`=(sqrtx-1)^2/(x-sqrtx)`
`=(sqrtx-1)/sqrtx`
`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`