\(A=\frac{2}{\sqrt{y+}\sqrt{x}}+\frac{1}{\sqrt{x}-\sqrt{y}}+\frac{3\sqrt{x}}{y-x}\)
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30 tháng 9 2015

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

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

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

\(=\frac{-\sqrt{y}}{x-y}\)

Tick cho mình nha bạn

10 tháng 8 2017

Ta có :

 Đặt A=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\left(\frac{x+y}{xy}\right).\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}.\left(\sqrt{x}+\sqrt{y}\right)^3}\right)\)

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

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

=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\frac{1}{xy}\)

=\(\frac{xy.\left(\sqrt{x}-\sqrt{y}\right)}{xy\sqrt{xy}}\)

=\(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)

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

=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{4-3}}\)

=\(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)

=> \(A^2=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)^2\)

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

           =\(4-2\sqrt{4-3}\)

           =\(4-2\)

           =\(2\)

=>\(A=\sqrt{2}\)

10 tháng 9 2019

các bn trả lời giúp mk câu hỏi đó vs 

10 tháng 9 2019

a/ \(\frac{\sqrt{a}-\left(\sqrt{a}\right)^2}{\sqrt{a}-1}\)

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

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

=\(-\sqrt{a}\)

15 tháng 8 2020

\(A=\frac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}+\frac{y}{\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)}-\frac{x+y}{\sqrt{xy}}\)

\(A=\frac{x\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)+y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)-\left(x+y\right)\left(y-x\right)}{\sqrt{xy}\left(y-x\right)}\)

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

\(A=\frac{\sqrt{xy}\left(x+y\right)}{\sqrt{xy}\left(y-x\right)}=\frac{y+x}{y-x}\)

KO CÓ GIÁ TRỊ y sao tính đây !!!!!!

CÒN      \(x=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)     nhé

14 tháng 10 2018

Q= [\(\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{y}+\sqrt{x}\right)}\)]\(:\frac{x-2\sqrt{xy}+y+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(Q=\left(\sqrt{x}+\sqrt{y}-\frac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right):\frac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)

\(Q=\frac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}.\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

\(Q=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

15 tháng 10 2018

phan 3 nua

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