a/ \(\sqrt{8\left(\sqrt{2}-\sqrt{3}\right)^2}=2\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)=2\sqrt{6}-4\)

b/ \(ab\sqrt{1+\frac{1}{a^2b^2}}=ab.\sqrt{\frac{a^2b^2+1}{a^2b^2}}=\sqrt{a^2b^2.\frac{a^2b^2+1}{a^2b^2}}=\sqrt{a^2b^2+1}\)

c/ \(\sqrt{\frac{a}{b^3}+\frac{a}{b^4}}=\sqrt{\frac{a}{b^3}\left(1+\frac{1}{b}\right)}=\frac{1}{b}.\sqrt{\frac{a}{b}\left(1+\frac{1}{b}\right)}\)

d/ \(\frac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\sqrt{a}\)

\(\sqrt{\frac{1}{600}}=\sqrt{\frac{6}{3600}}=\frac{\sqrt{6}}{\sqrt{3600}}=\frac{\sqrt{6}}{60}\)

\(\sqrt{\frac{11}{540}}=\sqrt{\frac{11}{36.15}}=\frac{1}{6}\sqrt{\frac{165}{15^2}}=\frac{1}{6}.\frac{\sqrt{165}}{15}=\frac{\sqrt{165}}{90}\)

\(\sqrt{\frac{3}{50}}=\sqrt{\frac{3}{25.2}}=\frac{1}{5}\sqrt{\frac{3}{2}}=\frac{1}{5}\sqrt{\frac{6}{4}}=\frac{1}{5}.\frac{\sqrt{6}}{2}=\frac{\sqrt{6}}{10}\)

\(\sqrt{\frac{5}{98}}=\sqrt{\frac{5}{49.2}}=\frac{1}{7}\sqrt{\frac{5}{2}}=\frac{1}{7}.\sqrt{\frac{10}{4}}=\frac{\sqrt{10}}{14}\)

\(\sqrt{\frac{\left(1-\sqrt{3}\right)^2}{27}}=\frac{\left|1-\sqrt{3}\right|}{\sqrt{9.3}}=\frac{\sqrt{3}-1}{3\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{9}\)

B4

a) \(\frac{9}{\sqrt{3}}=\frac{9\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\frac{9\sqrt{3}}{3}=3\sqrt{3}\)

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

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

d)\(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{1}=14\)

B3

a)\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\) \(đk:x\ge1\)

\(\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\sqrt{x-1}\cdot\left(\frac{1}{2}-\frac{9}{2}+3\right)=-17\)

\(\sqrt{x-1}\cdot\left(-1\right)=-17\)

\(\sqrt{x-1}=17\)

\(\left[{}\begin{matrix}x-1=289\left(tm\right)\\x-1=-289\left(ktm\right)\end{matrix}\right.\)

\(x=290\left(tm\right)\)

EM thử thôi, ko chắc đâu ạ:( Sai thì xin thông cảm cho ạ.

1) \(\sqrt{\frac{2}{3-\sqrt{5}}}=\sqrt{\frac{2\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}}=\sqrt{\frac{6+2\sqrt{5}}{4}}=\frac{\sqrt{6+2\sqrt{5}}}{2}\)

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

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

3) \(\sqrt{\frac{1}{a\left(1-\sqrt{3}\right)}}=\sqrt{\frac{1+\sqrt{3}}{a\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}}=\sqrt{\frac{1+\sqrt{3}}{a\left(1-3\right)}}=\sqrt{-\frac{1+\sqrt{3}}{2a}}\)

4) \(\sqrt{\frac{a}{4-2\sqrt{3}}}=\sqrt{\frac{a\left(4+2\sqrt{3}\right)}{\left(4-2\sqrt{3}\right)\left(4+2\sqrt{3}\right)}}=\sqrt{\frac{4a+2a\sqrt{3}}{16-12}}=\sqrt{\frac{4a+2a\sqrt{3}}{4}}=\frac{\sqrt{4a+2a\sqrt{3}}}{2}\)

a) \(A=\frac{3-\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}}=\sqrt{3}-1\)

b) \(B=\frac{\sqrt{6+2\sqrt{5}}}{\sqrt{5}+1}=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{5}+1}=\frac{\sqrt{5}+1}{\sqrt{5}+1}=1\)

c) \(C=\frac{2\sqrt{2}+\sqrt{6}}{4+\sqrt{12}}=\frac{2\sqrt{2}+\sqrt{6}}{4+2\sqrt{3}}=\frac{\left(2\sqrt{2}+\sqrt{6}\right)\left(4-2\sqrt{3}\right)}{\left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right)}=\frac{2\sqrt{2}}{4}=\frac{\sqrt{2}}{2}\)

d) \(D=\frac{\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{5+2\sqrt{6}}\left(\sqrt{2}-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=-\sqrt{5+2\sqrt{6}}\left(\sqrt{2}-\sqrt{3}\right)\)