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a/ \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}=2\sqrt{4.2.5\sqrt{4.3}}-2\sqrt{\sqrt{25.3}}-3\sqrt{5\sqrt{16.3}}\)
= \(2.2\sqrt{2.5.2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5.4\sqrt{3}}=4.2\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-3.2\sqrt{5\sqrt{3}}\)
= \(\sqrt{5\sqrt{3}}\left(8-2-6\right)=\sqrt{5\sqrt{3}}.0=0\)
b/ \(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}=2\sqrt{2.4\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{4.5\sqrt{3}}\)
= \(4\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=4\sqrt{2\sqrt{3}}-8\sqrt{5\sqrt{3}}\)
1,\(4\sqrt{5}+2\sqrt{5}-\sqrt{5}-15\sqrt{5}=-10\sqrt{5}\)
2,\(8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}=5\sqrt{5}\)
3,\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right):\sqrt{3}=33\)
4,\(7\sqrt{7a}+3\sqrt{7a}-2\sqrt{7a}=8\sqrt{7a}\)
5,\(-6\sqrt{a}-\sqrt{6a}+\sqrt{6a}=-6\sqrt{a}\)
6,\(8\sqrt{3}-12\sqrt{3}+5\sqrt{3}+2\sqrt{3}=3\sqrt{3}\)
a: \(=2\sqrt{\sqrt{3}}\cdot4\sqrt{5}-2\cdot\sqrt{\sqrt{3}}\cdot\sqrt{5}-3\cdot\sqrt{\sqrt{3}}\cdot2\sqrt{5}\)
\(=2\sqrt{\sqrt{3}}\left(4\sqrt{5}-\sqrt{5}-3\sqrt{5}\right)=0\)
b: \(=2\cdot2\sqrt{2}\cdot\sqrt{\sqrt{3}}-2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-3\cdot2\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
\(=2\sqrt{\sqrt{3}}\left(2\sqrt{2}-\sqrt{5}-3\sqrt{5}\right)=2\sqrt{\sqrt{3}}\cdot\left(2\sqrt{2}-4\sqrt{5}\right)\)
\(\left(\sqrt{12}+2\sqrt{27}-\sqrt{3}\right):\sqrt{3}\)
\(=\sqrt{12}:\sqrt{3}+2\sqrt{27}:\sqrt{3}-\sqrt{3}:\sqrt{3}\)
\(=\sqrt{4}+2\sqrt{9}-1\)
\(=2+6-1\)
\(=7\)
2) \(\left(4\sqrt{2}-\sqrt{8}+2\right).\sqrt{2-\sqrt{8}}\)
\(=\left(4\sqrt{2}-2\sqrt{2}+2\right).\sqrt{2-2\sqrt{2}}\)
\(=\left(2\sqrt{2}+2\right)^2.\left(\sqrt{2-2\sqrt{2}}\right)^2\)
\(=\left(8+4\right)\left(2-2\sqrt{2}\right)\)
\(=12.\left(2-2\sqrt{2}\right)\)
\(=24-24\sqrt{2}\)
\(=24\left(1-\sqrt{2}\right)\)
3) \(\sqrt{3}\left(2\sqrt{27}-\sqrt{75}+\frac{3}{2}\sqrt{12}\right)\)
\(=\sqrt{3}\left(2\sqrt{3^2.3}-\sqrt{5^2.3}+\frac{3}{2}\sqrt{2^2.3}\right)\)
\(=\sqrt{3}\left(6\sqrt{3}-5\sqrt{3}+3\sqrt{3}\right)\)
\(=\sqrt{3}.4\sqrt{3}\)
\(=12\)
\(a,=2\sqrt{3}-9\sqrt{3}+5\sqrt{3}=-2\sqrt{3}\)
\(b,=-1-4\sqrt{5}-2-12\sqrt{11}+10\sqrt{5}\)
\(=-3-4\sqrt{5}-12\sqrt{11}+10\sqrt{5}\)
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) \(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=\left(4\sqrt{2}-2\sqrt{5}-6\sqrt{5}\right).\sqrt{\sqrt{3}}\)
\(=\left(4\sqrt{2}-8\sqrt{5}\right).\sqrt{\sqrt{3}}\)
\(=\sqrt{\sqrt{3}}\left(\sqrt{2}-2\sqrt{5}\right)\)
Ta có: \(A=\sqrt{27}-2\sqrt{12}-\sqrt{75}\)
\(=3\sqrt{3}-2\cdot2\sqrt{3}-5\sqrt{3}\)
\(=-6\sqrt{3}\)
\(\Rightarrow A=3\sqrt{3}-2\cdot2\sqrt{3}-5\sqrt{3}=-4\sqrt{3}\)