a) \(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{5+2\sqrt{6}}\)

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

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

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

\(=2\sqrt{3}\)

b) \(\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-\sqrt{2}\)

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

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

\(=0\)

c) \(\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\cdot\left(2+\dfrac{5-3\sqrt{5}}{3-\sqrt{5}}\right)\)

\(=\left[2-\dfrac{\sqrt{5}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}\right]\cdot\left[2-\dfrac{\sqrt{5}\left(3-\sqrt{5}\right)}{3-\sqrt{5}}\right]\)

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

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

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

d) \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)

\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{6-4}-\dfrac{12\left(3+\sqrt{6}\right)}{9-6}\right]\left(\sqrt{6}+11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)

\(=6-121\)

\(=-115\)

a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)

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

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

\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)

c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)

a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)

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

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

b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)

\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)

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

\(=9+3\sqrt{5}-4\sqrt{5}+4\)

\(=13-\sqrt{5}\)

c) Ta có: \(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)

\(=\dfrac{10}{\sqrt{5}}+\dfrac{1}{5}\cdot5\sqrt{5}-2\cdot2\sqrt{5}\)

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

\(=-\sqrt{5}\)

e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)

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

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

f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

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

=3

e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)

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

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

f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

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

=3

a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)

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

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

b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)

\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)

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

\(=9+3\sqrt{5}-4\sqrt{5}+4\)

\(=13-\sqrt{5}\)

a) \(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}-\dfrac{6}{\sqrt{6}}=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}-\dfrac{6}{\sqrt{6}}\)

\(=\dfrac{1}{6\sqrt{6}}-\dfrac{6}{\sqrt{6}}=-\dfrac{35}{6\sqrt{6}}\)

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

\(=6+2\sqrt{30}+5+6-2\sqrt{30}+5=22\)

\(a,\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(1+\sqrt{5}\right)^2}\)

\(=\left|2-\sqrt{5}\right|-\left|1+\sqrt{5}\right|\)

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

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

\(=-3\)

\(b,\dfrac{3-5\sqrt{3}}{\sqrt{3}-5}+6\sqrt{\dfrac{4}{3}}\)

\(=\dfrac{\sqrt{3}\left(\sqrt{3}-5\right)}{\sqrt{3}-5}+6\cdot\dfrac{\sqrt{4}}{\sqrt{3}}\)

\(=\sqrt{3}+\dfrac{12}{\sqrt{3}}\)

\(=\sqrt{3}+\dfrac{12\sqrt{3}}{3}\)

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

\(=5\sqrt{3}\)

#\(Toru\)

\(\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(1+\sqrt{5}\right)^2}\\ =\left|2-\sqrt{5}\right|-\left|1+\sqrt{5}\right|\\ =\sqrt{5}-2-1-\sqrt{5}\\ =-2-1\\ =-3\)

\(\dfrac{3-5\sqrt{3}}{\sqrt{3}-5}+6\sqrt{\dfrac{4}{3}}\\ =\dfrac{\sqrt{3}\left(\sqrt{3}-5\right)}{\sqrt{3}-5}+4\sqrt{3}\\ =\sqrt{3}+4\sqrt{3}\\ =5\sqrt{3}\)

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

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

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

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

\(=2\left(3-\sqrt{3}\right)-3\sqrt{3}+\dfrac{3\sqrt{2}}{2}=6-5\sqrt{3}+\dfrac{3\sqrt{2}}{2}\) (câu này khả năng đề sai, dấu \(\sqrt{3}.\sqrt{2}\) ở mẫu cuối cùng là dấu trừ mới hợp lý)

\(C=\left(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)

\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}=\dfrac{3}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)^2}\)

Dấu giữa 2 dấu ngoặc là dấu chia sẽ hợp lý hơn

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

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

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

b: Sửa đề: \(\dfrac{7-2\sqrt{7}}{2-\sqrt{7}}+\dfrac{6}{\sqrt{7}+1}+\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)\)

\(=\dfrac{\sqrt{7}\left(\sqrt{7}-2\right)}{-\left(\sqrt{7}-2\right)}+\dfrac{6\left(\sqrt{7}-1\right)}{6}+18-12\)

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

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

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

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

\(=\left(2\sqrt{5}+4-10-4\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

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

\(=-20+2\sqrt{15}+\sqrt{15}-3-6\sqrt{5}+6\sqrt{3}\)

\(=-23+3\sqrt{15}-6\sqrt{5}+6\sqrt{3}\)