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

\(=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{4+4\sqrt{3}+3}{4-3}=\frac{7+4\sqrt{3}}{1}=7+4\sqrt{3}\)

a) \(\frac{3}{\sqrt{5}}=\frac{3\sqrt{5}}{\sqrt{5}.\sqrt{5}}=\frac{3\sqrt{5}}{5}\)

\(\frac{2\sqrt{3}}{\sqrt{2}}=\frac{2\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{2\sqrt{6}}{2}=\sqrt{6}\)

\(\frac{a}{\sqrt{b}}=\frac{a\sqrt{b}}{\sqrt{b}.\sqrt{b}}=\frac{a\sqrt{b}}{b}\)

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

bạn làm tương tự nha

câu c chắc là như này

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

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

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

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

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

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

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

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

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

b)\(\sqrt{4x^2-9}=2\sqrt{2x+3}\) \(đk:x\ge-\frac{3}{2}\)

\(\sqrt{\left(2x-3\right)\left(2x+3\right)}-2\sqrt{2x+3}=0\)

\(\sqrt{\left(2x+3\right)}\cdot\left(\sqrt{2x-3}-2\right)=0\)

\(\left[{}\begin{matrix}\sqrt{2x+3}=0\\\sqrt{2x-3}-2=0\end{matrix}\right.\left[{}\begin{matrix}2x+3=0\\\sqrt{2x-3}=2\end{matrix}\right.\left[{}\begin{matrix}x=-\frac{3}{2}\\2x-3=4\left(tm\right)\\2x-3=-4\left(ktm\right)\end{matrix}\right.\left[{}\begin{matrix}x=-\frac{3}{2}\left(tm\right)\\x=\frac{7}{2}\left(tm\right)\end{matrix}\right.\)

a.\(\frac{5}{\sqrt{10}}=\frac{5\sqrt{10}}{10}=\frac{\sqrt{10}}{2}\)

b. \(\frac{1}{3\sqrt{20}}=\frac{\sqrt{20}}{60}=\frac{2\sqrt{5}}{60}=\frac{\sqrt{5}}{30}\)

c. \(\frac{2\sqrt{2}+2}{5\sqrt{2}}=\frac{2\left(\sqrt{2}+1\right)}{5\sqrt{2}}=\frac{2\sqrt{2}\left(\sqrt{2}+1\right)}{10}=\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{5}\)

d.\(\frac{\sqrt{21}-\sqrt{7}}{1-\sqrt{3}}=\frac{\sqrt{7}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=\frac{-\sqrt{7}\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=-\sqrt{7}\)

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

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