a)\(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)

= \(2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)

= \(4\sqrt{5}\)

b) \(\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)

= \(\sqrt{3\left(5-2\sqrt{6}\right)}-\sqrt{33-12\sqrt{6}}\)

= \(\sqrt{3\left(5-2\sqrt{6}\right)}-\sqrt{3\left(11-4\sqrt{6}\right)}\)

\(a,2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)

\(=2\sqrt{5}-\sqrt{5^2.5}-\sqrt{4^2.5}+\sqrt{11^2.5}\)

\(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)

\(=4\sqrt{5}\)

\(b,\sqrt{15-\sqrt{216}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)

\(=|3-\sqrt{6}|+|3-2\sqrt{6}|\)

\(=3-\sqrt{6}+2\sqrt{6}-3\)

\(=\sqrt{6}\)

\(x+2y=\sqrt{\left(\frac{1}{\sqrt{2}}.\sqrt{2}x+\frac{2}{\sqrt{3}}.\sqrt{3}y\right)^2}\le\sqrt{\left(\frac{1}{2}+\frac{4}{3}\right)\left(2x^2+3y^2\right)}=\sqrt{\frac{22}{3}}\)

\(3=x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\)\(\Leftrightarrow\)\(x+y+z\le3\)

\(x^3+y^3+z^3=\frac{x^4}{x}+\frac{y^4}{y}+\frac{z^4}{z}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x+y+z}\ge\frac{3^2}{3}=3\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=1\)

Ta có:\(x;y;z>0\Leftrightarrow x^3;y^3;z^3\ge0\Leftrightarrow x^3\ge x^2;y^3\ge y^2;z^3\ge z^2\)

\(\Leftrightarrow x^3+y^3+z^3\ge x^2+y^2+z^2hay:x^3+y^3+z^3\ge3\)

a) \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\)

= \(\frac{\left(10+2\sqrt{10}\right)\left(1-\sqrt{5}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(1-\sqrt{5}\right)}+\frac{8\left(\sqrt{5}+\sqrt{2}\right)}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)

= \(\frac{\left(10+2\sqrt{10}\right)\left(1-\sqrt{5}\right)+8\left(\sqrt{5}+\sqrt{2}\right)}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)

= \(\frac{10-2\sqrt{5}+2\sqrt{10}-2\sqrt{2}}{\sqrt{5}+\sqrt{2}-5-\sqrt{10}}\)

= \(\frac{2\left(5-\sqrt{5}+\sqrt{10}-\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}-5-\sqrt{10}}\)

= -2

b); c); d) làm tương tự

a/\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)

\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)

b/ \(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)\)

\(=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)

\(=2\sqrt{3}\cdot6\sqrt{3}=2\cdot6\cdot3=36\)

c/ \(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)\)

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

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

\(=2+2\sqrt{3}\)

d/ \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{13-4\sqrt{10}}-\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{8-4\sqrt{10}+5}-\sqrt{45+12\sqrt{10}+8}\)

\(=\sqrt{\left(2\sqrt{2}\right)^2-2\cdot2\sqrt{2\cdot5}+\left(\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\cdot2\sqrt{5\cdot2}+\left(2\sqrt{2}\right)^2}\)

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

\(=2\sqrt{2}-\sqrt{5}-3\sqrt{5}-2\sqrt{2}\)

\(=-4\sqrt{5}\)

#)Giải :

 \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)