Question

[Tashigi]

Rút gọn

A=\(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)\)

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

\(D=\sqrt{4+\sqrt{8}}\sqrt{2-\sqrt{2+\sqrt{2}}}\sqrt{2+\sqrt{2+\sqrt{2}}}\)

Answer 1

[๖ACE✪Hoàngミ★Việtツ]

\(A=\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)\)

\(A=\left(\frac{5+2\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}+\frac{5-2\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\left(15+2\sqrt{6}\right)\)

\(A=\left(\frac{5+2\sqrt{6}+5-2\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\left(15+2\sqrt{6}\right)\)

\(A=\left(\frac{10}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\left(15+2\sqrt{6}\right)\)

\(A=\left(\frac{10}{25-24}\right)\left(15+2\sqrt{6}\right)\)

\(A=10\left(15+2\sqrt{6}\right)\)

\(A=150+20\sqrt{6}\)