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

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

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

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

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

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

24 tháng 7 2017

a, dk \(x\ge0.x\ne1\)

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

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

phan b,c ban tu lam not nhe dai lam mk ko lam dau  mk co vc ban rui

2 tháng 8 2019

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

\(P=\left[-\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\left(-\frac{\sqrt{x}}{2}+\frac{1}{2\sqrt{x}}\right)^2\)

\(P=\left[-\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\left(\frac{1}{4x}+\frac{1}{4}-\frac{1}{2}\right)\)

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

\(P=-\frac{4.\frac{x^2-2x+1}{4x}.\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\)

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

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

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

25 tháng 8 2020

ĐKXĐ: \(x\ge1\); x khác 2; 3

Ta có: 

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

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

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

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

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

29 tháng 6 2019

ĐK : x>0, x khác 1

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

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

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

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

11 tháng 7 2018

Bài 1:

a)  \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)

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

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

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

b)   \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)

\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)

\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)

c)  ĐK:  \(a\ge0;a\ne1\)

  \(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)

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

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

\(=1-a+a=1\)

12 tháng 6 2019

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

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

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

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

12 tháng 6 2019

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

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

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

\(=\frac{x^2-x\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{2\sqrt{x}}\)

\(=\frac{x^2-2x\sqrt{x}-2x-\sqrt{x}}{2\sqrt{x}}\)