Q=\(\frac{a}{\sqrt{a^2-b^2}}-\frac{\sqrt{a^2-b^2}+a}{\sqrt{a^2-b^2}}\times\frac{a-\sqrt{a^2-b^2}}{b}\)

Q=\(\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-a^2+b^2}{b\sqrt{a^2-b^2}}\)

Q= \(\frac{a+b}{\sqrt{a^2-b^2}}\)

Q=\(\frac{\sqrt{a+b}}{\sqrt{a-b}}\)

C= \(\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)          -  \(\frac{2}{\sqrt{ab}}\); \(\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2\)

= \(\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)-   \(\frac{2}{\sqrt{ab}}\).: \(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{ab}\)

= \(\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)-\(\frac{2\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

= \(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)

=1

#mã mã#

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

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

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

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

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

\(A=a-\sqrt{a}\)

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

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

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

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

\(=\frac{-2.2\sqrt{x}}{2}\)

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

Thank bạn bài vừa rồi đã k cho mk^^

Bạn ơi bạn đã giải được bài 1 chưa vậy? 

Bạn xem có nhầm đề bài không phải là\(\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}\) thế mới sử dụng đc trục căn thức ở mẫu