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

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

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+b\right)}-\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{4b\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2-4b\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}\right)-4\sqrt{a}b-4b\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{2\sqrt{a}.2\sqrt{b}-4\sqrt{a}b-4b\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{4\sqrt{a}\sqrt{b}-4\sqrt{a}b-4b\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{4\sqrt{a}\sqrt{b}\left(1-\sqrt{b}-b\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{2\sqrt{a}\sqrt{b}\left(1-\sqrt{b}-b\right)}{a-b}\)

Đề sai???Phân số thứ 3 nghi là a-b chứ ko phải căn a - căn b????????

ta tính VT ra xong rồi nói VT = VP

Biến đổi Vế trái ta có :

 \(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{3}\left(2-\sqrt{2}\right)}{\sqrt{2}\left(2-\sqrt{2}\right)}-\frac{3\sqrt{24}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

\(\left(\frac{\sqrt{3}}{\sqrt{2}}-2\sqrt{6}\right)\cdot\frac{1}{\sqrt{6}}=\frac{\sqrt{3}}{\sqrt{2}}\cdot\frac{1}{\sqrt{6}}-2\sqrt{6}\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{1}{2}-2=-1,5=VP\)  ( ĐPCM) 

Biến đổi vế trái :

\(VT=\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{2}.\sqrt{2}.\sqrt{3}-\sqrt{6}}{\sqrt{2^2.2}-2}-\frac{\sqrt{6^2.6}}{3}\right).\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{2}.\sqrt{6}-\sqrt{6}}{2\sqrt{2}-2}-\frac{6\sqrt{6}}{3}\right).\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}.\left(\sqrt{2}-1\right)}{2.\left(\sqrt{2}-1\right)}-2\sqrt{6}\right).\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}}{2}-2\sqrt{6}\right).\frac{1}{\sqrt{6}}\)

\(=\sqrt{6}.\left(\frac{1}{2}-2\right).\frac{1}{\sqrt{6}}=-\frac{3}{2}=-1,5=VP\left(đpcm\right)\)

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

\(=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{4b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)

ò, Linh ơi, mình nghĩ bạn làm đúng nhưng mà chỗ dấu ''='' thứ nhất bạn ghi ''4b'' nhưng bước đó bạn phải ghi là ''2b'' tại bước đó chưa có quy đồng, quy đồng mới thành 4b do mẫu chung là \(2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\), chắc bạn hiểu, cảm ơn bạn nhiều nha!