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25 tháng 8 2020

ĐKXĐ: \(x\ge1\); x khác 2; 3

Ta có: 

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

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

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

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

=> \(P=\left(\sqrt{x}-\sqrt{2}\right).\frac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}=\frac{2-\sqrt{x}}{\sqrt{x}}\)

26 tháng 11 2017

\(P=\frac{2x+2}{\sqrt{x}}+\frac{x\sqrt{x}+1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\)

\(P=\frac{2x+2}{\sqrt{x}}+\frac{\sqrt{x^3}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x^3}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

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

\(P=\frac{2x+2}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\)

\(P=\frac{2x+2-x+\sqrt{x}-1-x+\sqrt{x}-1}{\sqrt{x}}\)

\(P=\frac{2\sqrt{x}}{\sqrt{x}}\)

\(P=2\)

vậy  \(P=2\)

3 tháng 6 2019

ĐK:x>1

M=\(\frac{x-1}{2x}\) .\(\frac{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}\right)\left(\sqrt{x}+1\right)}{x-1}\)

=\(\frac{x-1}{2x}\).\(\frac{x\sqrt{x}-x-x+\sqrt{x}-x\sqrt{x}-x-x-\sqrt{x}}{x-1}\)=\(\frac{x-1}{2x}\).\(\frac{-4x}{x-1}\)=-2

Vậy M=-2