\(\sqrt{11+6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)

Mik ko hiểu đề lắm

\(\sqrt{11+6\sqrt{2}}=3+\sqrt{2}\)

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

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

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

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

\(a,=-2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\sqrt{5}=-18\sqrt{5}\)

\(b,=2\sqrt{3}-5\sqrt{3}+4\sqrt{3}-7\sqrt{3}=-6\sqrt{3}\)

\(c,=3\sqrt{3}+7\sqrt{3}-9\sqrt{3}+11\sqrt{3}=12\sqrt{3}\)

a) Ta có: \(-\sqrt{20}+3\sqrt{45}-6\sqrt{80}-\dfrac{1}{5}\sqrt{125}\)

\(=-2\sqrt{5}+9\sqrt{5}-24\sqrt{5}-\dfrac{1}{5}\cdot5\sqrt{5}\)

\(=-17\sqrt{5}-\sqrt{5}=-18\sqrt{5}\)

b) Ta có: \(2\sqrt{3}-\sqrt{75}+2\sqrt{12}-\sqrt{147}\)

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

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

\(A=\sqrt{64a^2}\cdot2a=\sqrt{\left(8a\right)^2}\cdot2a=\left|8a\right|\cdot2a\)

Với a < 0 A = 8a.(-2a) = -16a²

Với a ≥ 0 A = 8a.2a = 16a²

\(B=3\sqrt{9a^6}-6a^3=3\sqrt{\left(3a^3\right)^2}-6a^3=9\left|a^3\right|-6a^3\)