a) \(\dfrac{2\sqrt{125}-3\sqrt{5}-\sqrt{180}}{-\sqrt{5}}+\sqrt{8}=\dfrac{2\sqrt{25.5}-3\sqrt{5}-\sqrt{36.5}}{-\sqrt{5}}+\sqrt{8}\)

\(=\dfrac{10\sqrt{5}-3\sqrt{5}-6\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=\dfrac{\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=2\sqrt{2}-1\)

b) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}\)

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

c) \(\sqrt{48}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}=\sqrt{16.3}-2\sqrt{9.\dfrac{1}{3}}+\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}\)

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

d) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right).\left(\sqrt{5}-\sqrt{2}\right)\)

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

\(a,=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}\)

\(=\sqrt{2}\left(3-12+8-5\right)=-6\sqrt{2}\)

\(b,=\left|\sqrt{2}-\sqrt{3}\right|+3\sqrt{2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}=\sqrt{3}+2\sqrt{2}\)

\(c,=\sqrt{5}+\sqrt{5}+\dfrac{5}{\sqrt{5}}-1=3\sqrt{5}-1\)

\(d,=\sqrt{3-2.2\sqrt{3}+4}+\sqrt{\left(1+\sqrt{3}\right)^2}\)

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

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

a) \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}=3\sqrt{2}-4\sqrt{9.2}+2\sqrt{16.2}-\sqrt{25.2}\)

\(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}=-6\sqrt{2}\)

b) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}\)

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

c) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}=\sqrt{25.\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{9.5}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}\)

\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1=3\sqrt{5}-1\)

d) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}+\left|\sqrt{3}+1\right|\)

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

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

\(=\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{24-2\cdot2\sqrt{6}\cdot3+9}\)

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

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

b: \(\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)

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

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

\(=3+\sqrt{5}+3-\sqrt{5}=6\)

c: \(\dfrac{3}{2\sqrt{3}+3}+\dfrac{3}{2\sqrt{3}-3}\)

\(=\dfrac{3\left(2\sqrt{3}-3\right)+3\left(2\sqrt{3}+3\right)}{12-9}\)

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

d: \(\sqrt{\left(\sqrt{3}+4\right)\cdot\sqrt{19-8\sqrt{3}}+3}\)

\(=\sqrt{\left(4+\sqrt{3}\right)\cdot\sqrt{\left(4-\sqrt{3}\right)^2}+3}\)

\(=\sqrt{\left(4+\sqrt{3}\right)\cdot\left(4-\sqrt{3}\right)+3}\)

\(=\sqrt{16-3+3}=\sqrt{16}=4\)

e: \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)

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

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

a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)

b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)

c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)

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

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

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

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

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

\(=\sqrt{\left(2-\sqrt{3}\right)\left(8+4\sqrt{3}\right)}\)

\(=\sqrt{\left(2-\sqrt{3}\right)\cdot4\left(2+\sqrt{3}\right)}\)

\(=\sqrt{\left(4-3\right)\cdot4}\)

\(=\sqrt{1\cdot4}\)

\(=\sqrt{4}\)

\(=2\)

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

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

\(=2\sqrt{2}+6+3\sqrt{2}+1-\left(5\sqrt{2}-7\right)\)

\(=2\sqrt{2}+6+3\sqrt{2}+1-5\sqrt{2}+7\)

\(=0+14\)

\(=14\)

c) \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)

dài quá ==' cả d, e, f nữa ==' có j rảnh lm cho nhé :D

a. \(\sqrt{48}-2\sqrt{32}-\sqrt{75}+3\sqrt{50}\) = \(4\sqrt{3}-2.4\sqrt{2}-5\sqrt{3}+3.5\sqrt{2}\)

= \(4\sqrt{3}-8\sqrt{2}-5\sqrt{3}+15\sqrt{2}\)  = \(-\sqrt{3}+7\sqrt{2}\)

b. \(\sqrt{20}-15\sqrt{\dfrac{1}{5}}+\sqrt{\left(1-\sqrt{5}\right)^2}\) = \(2\sqrt{5}-3.5.\sqrt{\dfrac{1}{5}}+\left|1-\sqrt{5}\right|\)

= \(2\sqrt{5}-3\sqrt{25.\dfrac{1}{5}}+\sqrt{5}-1\) =  \(2\sqrt{5}-3\sqrt{5}+\sqrt{5}-1\) = \(-1\)

c. \(\dfrac{3}{3+2\sqrt{3}}+\dfrac{3}{3-2\sqrt{3}}\) = \(\dfrac{3\left(3-2\sqrt{3}\right)+3\left(3+2\sqrt{3}\right)}{\left(3+2\sqrt{3}\right)\left(3-2\sqrt{3}\right)}\) 

= \(\dfrac{9-6\sqrt{3}+9+6\sqrt{3}}{\left(3+2\sqrt{3}\right)\left(3-2\sqrt{3}\right)}\) = \(\dfrac{18}{9-12}=\dfrac{18}{-3}=-6\)

\(B=\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

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

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

\(B=\sqrt{5}+2+\sqrt{5}-2\)

\(B=2\sqrt{5}\)

\(A=\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}\)

\(A=\left(\dfrac{\sqrt{12}-\sqrt{6}}{2\sqrt{2}-2}-\dfrac{6\sqrt{6}}{3}\right).\dfrac{1}{\sqrt{6}}\)

\(A=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}\)

\(A=\left(\sqrt{6}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}\)

\(A=-\sqrt{6}.\dfrac{1}{\sqrt{6}}\)

\(A=-1\)

a) Ta có: \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)

\(=\dfrac{-2\left(\sqrt{3}-\sqrt{8}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{6}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)

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

b) Ta có: \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1-\sqrt{2}-\sqrt{3}\right)\)

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

\(=1-5-2\sqrt{6}\)

\(=-4-2\sqrt{6}\)