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Ta có : \(\frac{20\sqrt{300}-15\sqrt{675}+5\sqrt{75}}{\sqrt{15}}=\frac{\sqrt{15}\left(20\sqrt{20}-15\sqrt{45}+5\sqrt{5}\right)}{\sqrt{15}}\)
\(=20\sqrt{20}-15\sqrt{45}+5\sqrt{5}=40\sqrt{5}-45\sqrt{5}+5\sqrt{5}=0\)
a,
\(\sqrt{4-2\sqrt{3}}-\sqrt{3}\\ =\sqrt{3-2\cdot1\cdot\sqrt{3}+1}-\sqrt{3}\\ =\sqrt{\left(\sqrt{3}\right)^2-2\cdot1\cdot\sqrt{3}+1^2}-\sqrt{3}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\\ =\sqrt{3}-1-\sqrt{3}\\ =-1\)
b,
\(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\\ =\sqrt{9+2\cdot3\cdot\sqrt{2}+2}-3+\sqrt{2}\\ =\sqrt{3^2+2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-3+\sqrt{2}\\ =\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}\\ =3+\sqrt{2}-3+\sqrt{2}\\ =2\sqrt{2}\)
c,
\(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}\\ =\sqrt{5+2\cdot\sqrt{2\cdot5}+2}-\sqrt{5-2\cdot\sqrt{2\cdot5}+2}\\ =\sqrt{\left(\sqrt{5}\right)^2+2\cdot\sqrt{2}\cdot\sqrt{5}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{2}\cdot\sqrt{5}+\left(\sqrt{2}\right)^2}\\ =\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}\\ =2\sqrt{2}\)
d,
\(\left(20\sqrt{300}-15\sqrt{675}+5\sqrt{75}\right):\sqrt{15}\\ =\left(20\cdot\sqrt{20}\cdot\sqrt{15}-15\cdot\sqrt{45}\cdot\sqrt{15}+5\cdot\sqrt{5}\cdot\sqrt{15}\right):\sqrt{15}\\ =\left(20\cdot2\cdot\sqrt{5}\cdot\sqrt{15}-15\cdot3\cdot\sqrt{5}\cdot\sqrt{15}+5\cdot\sqrt{5}\cdot\sqrt{15}\right):\sqrt{15}\\ =\sqrt{15}\cdot\left(20\cdot2\cdot\sqrt{5}-15\cdot3\cdot\sqrt{5}+5\cdot\sqrt{5}\right):\sqrt{15}\\ =20\cdot2\cdot\sqrt{5}-15\cdot3\cdot\sqrt{5}+5\cdot\sqrt{5}\\ =40\sqrt{5}-45\sqrt{5}+5\sqrt{5}\\ =0\)
a,
\(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)3\sqrt{6}\)
\(=6\sqrt{6}^2-12\sqrt{9.2}+15\sqrt{4.3}-\dfrac{3}{4}\sqrt{16.3}\)
= \(6.6-12.3\sqrt{2}+15.2\sqrt{3}-\dfrac{3}{4}.4\sqrt{3}\)
\(=36-36\sqrt{2}+30\sqrt{3}-3\sqrt{3}\)
\(=36-36\sqrt{2}+27\sqrt{3}\)
\(A=\sqrt{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}\)
=> \(A=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{4+2\sqrt{3}}\)
=> \(A=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
=> \(A=\left(\sqrt{3}+1\right)^2\left(\sqrt{3}-2\right)\)
=> \(A=\left(4+2\sqrt{3}\right)\left(\sqrt{3}-2\right)\)
=> \(A=4\sqrt{3}-8+6-4\sqrt{3}\)
=> \(A=-8+6=-2\)
VẬY \(A=-2\)
\(B=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{2}.\sqrt{4-\sqrt{15}}\)
=> \(B=\sqrt{8-2\sqrt{15}}\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\)
=> \(B=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\left(4+\sqrt{15}\right)\)
=> \(B=\left(\sqrt{5}-\sqrt{3}\right)^2\left(4+\sqrt{15}\right)\)
=> \(B=\left(8-2\sqrt{15}\right)\left(4+\sqrt{15}\right)\)
=> \(B=32+8\sqrt{15}-8\sqrt{15}-30\)
=> \(B=2\)
VẬY \(B=2\)
\(20\sqrt{300}-15\sqrt{675}+5\sqrt{75}\)
\(=20\sqrt{10^2.3}-15\sqrt{15^2.3}+5\sqrt{5^2.3}\)
\(=20.10\sqrt{3}-15.15\sqrt{3}+5.5\sqrt{3}\)
\(=200\sqrt{3}-225\sqrt{3}+25\sqrt{3}=0\)