K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

10 tháng 8 2017

Ta có :

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

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

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

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

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

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

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

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

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

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

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

           =\(4-2\sqrt{4-3}\)

           =\(4-2\)

           =\(2\)

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

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

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

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

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

\(=\dfrac{\sqrt{xy}\left(x+y\right)\cdot\left(\sqrt{x}-\sqrt{y}\right)}{-\sqrt{xy}\left(x+y\right)}=-\sqrt{x}+\sqrt{y}\)(1)

Khi x=3 và \(y=4+2\sqrt{3}\) vào (1), ta được:

\(=-\sqrt{3}+\sqrt{4+2\sqrt{3}}=-\sqrt{3}+\sqrt{3}+1=1\)