a: \(=2\sqrt{20\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\cdot\sqrt{20\sqrt{3}}\)

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

b: \(=2\sqrt{5\sqrt{3}}-4\sqrt{2\sqrt{3}}-6\sqrt{5\sqrt{3}}=-4\sqrt{5\sqrt{3}}-4\sqrt{2\sqrt{3}}\)

cảm ơn anh ạ 

a) \(E=2\sqrt{40\sqrt{12}}+3\sqrt{5\sqrt{48}}-2\sqrt{\sqrt{75}}-4\sqrt{15\sqrt{27}}.\)

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

  \(=0\)

b) \(F=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}.\)

Vì \(=\frac{5}{12}-\frac{1}{\sqrt{6}}=\frac{5-2\sqrt{6}}{12}=\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}\)

\(\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}=\frac{2\sqrt{3}+\sqrt{2}}{6}\)

Nên \(F=\frac{2\sqrt{3}+\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}\)

Ta có: \(\dfrac{2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+\sqrt{2}}\)

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

\(=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{2}\left(\sqrt{6}+1\right)}\)

\(=\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)\left(\sqrt{6}+1\right)-\sqrt{15}-\sqrt{5}}{\sqrt{2}\left(\sqrt{6}+1\right)}\)

\(=\dfrac{\sqrt{2}\left(3\sqrt{6}+3+\sqrt{30}+\sqrt{5}\right)-\sqrt{15}-\sqrt{5}}{\sqrt{2}\left(\sqrt{6}+1\right)}\)

\(=\dfrac{6\sqrt{3}+3\sqrt{2}+2\sqrt{15}+\sqrt{10}-\sqrt{15}-\sqrt{5}}{\sqrt{2}\left(\sqrt{6}+1\right)}\)

\(=\dfrac{6\sqrt{3}+3\sqrt{2}+\sqrt{15}+\sqrt{10}-\sqrt{5}}{ }\)

Đề sai rồi bạn

Ta có: \(\dfrac{2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)

\(=\dfrac{\left(6+2\sqrt{5}\right)\sqrt{6-2\sqrt{5}}}{\sqrt{20}-2}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}+1\right)}\)

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

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

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

\(=2\sqrt{80\sqrt{3}}+3\sqrt{20\sqrt{3}}-2\sqrt{5\sqrt{3}}-4\sqrt{45\sqrt{3}}\)

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

=0

a) \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)

\(=2\sqrt{40.2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5.4\sqrt{3}}\)

\(=\left(2\sqrt{80}-2\sqrt{5}-3\sqrt{20}\right).\sqrt{\sqrt{3}}\)

\(=\left(8\sqrt{5}-2\sqrt{5}-6\sqrt{5}\right).\sqrt{\sqrt{3}}=0\)

b) \(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)

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

\(=\left(4\sqrt{2}-8\sqrt{5}\right).\sqrt{\sqrt{3}}\)

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

sai rồi

ở đáp án còn số 4 ở đầu nữa

Xét tử \(2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}\)

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

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

\(=2\sqrt{2+\sqrt{3}}=\frac{2\sqrt{4+2\sqrt{3}}}{\sqrt{2}}=\frac{2\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}=\frac{2\left(\sqrt{3}+1\right)}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)

Suy ra VT = VP = 1

Đáp án là : VT = VP = 1

vhuv bn hoc gioi tk mk nha

cac ban xem the nay co dung ko nha:

\(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)

\(=2\sqrt{40\sqrt{3.2^2}}-2\sqrt{\sqrt{3.5^2}}-3\sqrt{5\sqrt{3.4^2}}\)

\(=2\sqrt{40.2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5.4\sqrt{3}}\)

\(=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)

\(=2\sqrt{5.4^2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{2^2.5\sqrt{3}}\)

\(=2.4\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-3.2\sqrt{5\sqrt{3}}\)

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

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

\(=0\)

moi nguoi cho xin y kien