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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) \(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=\left(4\sqrt{2}-2\sqrt{5}-6\sqrt{5}\right).\sqrt{\sqrt{3}}\)
\(=\left(4\sqrt{2}-8\sqrt{5}\right).\sqrt{\sqrt{3}}\)
\(=\sqrt{\sqrt{3}}\left(\sqrt{2}-2\sqrt{5}\right)\)
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
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
Bạn áp dụng hằng đẳng thức (a+b+c)^2= a^2+b^2+c^2+2(ab+ac+bc)
Biến đổi vế trái ta có:
\(\sqrt{3+\sqrt{5}-\sqrt{13+4\sqrt{3}}}=\sqrt{3+\sqrt{5}-\sqrt{\left(2\sqrt{3}+1\right)^2}}=\sqrt{3+\sqrt{5}-2\sqrt{3}-1}=\sqrt{2+\sqrt{5}-2\sqrt{3}}\)
Đề sai
\(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}=2.\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=8\sqrt{5}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-6\sqrt{5}.\sqrt{\sqrt{3}}=0\)