Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
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: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
Ta có :
Đặt A=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\left(\frac{x+y}{xy}\right).\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}.\left(\sqrt{x}+\sqrt{y}\right)^3}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{x+y}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\frac{1}{xy}\)
=\(\frac{xy.\left(\sqrt{x}-\sqrt{y}\right)}{xy\sqrt{xy}}\)
=\(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{4-3}}\)
=\(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
=> \(A^2=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)^2\)
=\(2-\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+2+\sqrt{3}\)
=\(4-2\sqrt{4-3}\)
=\(4-2\)
=\(2\)
=>\(A=\sqrt{2}\)