Ta có: \(\frac{2+2\sqrt5}{3-\sqrt5}\cdot\sqrt{\frac{24-8\sqrt5}{3+3\sqrt5}}\)

\(=\sqrt{\frac{8\left(3-\sqrt5\right)}{3\left(\sqrt5+1\right)}\cdot\left(\frac{2+2\sqrt5}{3-\sqrt5}\right)^2}\)

\(=\sqrt{\frac{8\left(3-\sqrt5\right)}{3\left(\sqrt5+1\right)}\cdot\frac{4\left(\sqrt5+1\right)^2}{\left(3-\sqrt5\right)^2}}=\sqrt{\frac{8\cdot4\cdot\left(\sqrt5+1\right)}{3\left(3-\sqrt5\right)}}=\sqrt{\frac{32\left(\sqrt5+1\right)}{3\left(3-\sqrt5\right)}}\)

3√5 = √(32.5)=√45

\(a\cdot\sqrt{\dfrac{-15}{a}}=\sqrt{\dfrac{-15a^2}{a}}=\sqrt{-15a}\)

\(a\sqrt{\dfrac{-15}{a}}=\sqrt{a^2.\dfrac{-15}{a}}=\sqrt{\dfrac{-15a^2}{a}}=\sqrt{-15a}\left(a< 0\right)=\sqrt{15a}\)

1: \(3\sqrt8-5\sqrt{18}\)

\(=3\cdot2\sqrt2-5\cdot3\sqrt2\)

\(=6\sqrt2-15\sqrt2=-9\sqrt2\)

2:

\(7\sqrt3=\sqrt{7^2\cdot3}=\sqrt{147}\)

mà 147>141

nên \(7\sqrt3>\sqrt{141}\)

3: \(\sqrt{\frac{5}{27}}=\sqrt{\frac{5}{9\cdot3}}=\sqrt{\frac{15}{81}}=\frac{\sqrt{15}}{9}\)

\(\sqrt{\frac{11}{64}}=\frac{\sqrt{11}}{\sqrt{64}}=\frac{\sqrt{11}}{8}\)

\(\left(x-5\right)\sqrt{\frac{3}{25-x^2}}=\sqrt{\left(x-5\right)^2}\sqrt{\frac{3}{\left(5-x\right)\left(x+5\right)}}=\sqrt{\left(5-x\right)^2.\frac{3}{\left(5-x\right)\left(x+5\right)}}=\sqrt{\frac{3\left(5-x\right)}{x+5}}\)

thanks 

1,2√5 = √(1,22.5)= √7,2

\(x\sqrt{\dfrac{2}{x}}=\sqrt{x^2\cdot\dfrac{2}{x}}=\sqrt{2x}\)

\(x\sqrt{\dfrac{2}{5}}=\sqrt{\dfrac{2}{5}\cdot x^2}=\sqrt{\dfrac{2x^2}{5}}\)

\(\left(x-5\right)\cdot\sqrt{\dfrac{x}{25-x^2}}=\sqrt{\left(x-5\right)^2\cdot\dfrac{x}{-\left(x-5\right)\left(x+5\right)}}=\sqrt{-\dfrac{x\left(x-5\right)}{x+5}}\)

\(x\sqrt{\dfrac{7}{x^2}}=\sqrt{x^2\cdot\dfrac{7}{x^2}}=\sqrt{7}\)