a) Kết quả rút gọn xấu (+dài) nữa. (có thể đề sai)

b) 

\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left[\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right].\left(\sqrt{7}-\sqrt{5}\right)\)

\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-\left(7-5\right)=-2\)

c) \(\frac{\sqrt{5-2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}=\frac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)

\(=\frac{\sqrt{3}-\sqrt{2}+\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\frac{\left(\sqrt{5}-\sqrt{2}\right)^2}{3}\)

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

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

Thêm câu này hộ tớ nx nhé !
e) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right).\left(\sqrt{2}-3\sqrt{0.4}\right)\)

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

\(=\left(\frac{\sqrt{12}-\sqrt{6}}{2\left(\sqrt{2}-1\right)}-\frac{6\sqrt{6}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

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

\(=\left(\frac{\sqrt{6}}{2}-\frac{4\sqrt{6}}{2}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{\sqrt{6}-4\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)

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

\(=-\frac{3}{2}\)

\(b,\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

\(=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

\(=\left(-\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

\(=-\left(7-5\right)\)

\(=-2\)

* \(\frac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\frac{6\sqrt{2}-4}{3-\sqrt{2}}\)\(=\frac{\left(\sqrt{8}-\sqrt{7}\right)}{\left(\sqrt{8}+\sqrt{7}\right)\left(\sqrt{8}-\sqrt{7}\right)}+\sqrt{25.7}-\frac{2\sqrt{2}\left(3-\sqrt{2}\right)}{3-\sqrt{2}}\)

\(=\sqrt{8}-\sqrt{7}+5\sqrt{7}-2\sqrt{2}=2\sqrt{2}+4\sqrt{7}-2\sqrt{2}=4\sqrt{7}\)

** \(\frac{\sqrt{6-\sqrt{11}}}{\sqrt{22}-\sqrt{2}}+\frac{6}{\sqrt{2}}-\frac{3}{\sqrt{2}+1}\)\(=\frac{\sqrt{2}\sqrt{6-\sqrt{11}}}{\sqrt{2}\left(\sqrt{22}-\sqrt{2}\right)}+\frac{6\sqrt{2}}{2}-\frac{3\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\)

\(=\frac{\sqrt{12-2\sqrt{11}}}{2\sqrt{11}-2}+3\sqrt{2}-\frac{3\sqrt{2}-3}{1}\)\(=\frac{\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}+1^2}}{2\left(\sqrt{11}-1\right)}+3\sqrt{2}-3\sqrt{2}+3\)

\(=\frac{\sqrt{11}-1}{2\left(\sqrt{11}-1\right)}+3=\frac{1}{2}+3=\frac{7}{2}\).

Cảm ơn bạn :)

Nhân cả tử và mẫu với biểu thức liên hợp của mẫu (câu a mẫu cuối kì kì)

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

\(B=-\left(\sqrt{1}+\sqrt{2}-\sqrt{2}-\sqrt{3}+\sqrt{3}+\sqrt{4}-...+\sqrt{7}+\sqrt{8}-\sqrt{8}-\sqrt{9}\right)\)

\(B=-\left(\sqrt{1}-\sqrt{9}\right)=2\)