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\(\left(\sqrt{12}+2\sqrt{27}-\sqrt{3}\right):\sqrt{3}\)
\(=\sqrt{12}:\sqrt{3}+2\sqrt{27}:\sqrt{3}-\sqrt{3}:\sqrt{3}\)
\(=\sqrt{4}+2\sqrt{9}-1\)
\(=2+6-1\)
\(=7\)
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) \(\sqrt{3}\left(2\sqrt{27}-\sqrt{75}+\frac{3}{2}\sqrt{12}\right)\)
\(=\sqrt{3}\left(2\sqrt{3^2.3}-\sqrt{5^2.3}+\frac{3}{2}\sqrt{2^2.3}\right)\)
\(=\sqrt{3}\left(6\sqrt{3}-5\sqrt{3}+3\sqrt{3}\right)\)
\(=\sqrt{3}.4\sqrt{3}\)
\(=12\)
1.\(D=\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)\(=2\sqrt{3}-10\sqrt{3}-\sqrt{3}+\frac{10\sqrt{3}}{3}\)\(=\frac{-17\sqrt{3}}{3}\)
2.\(A=27-\left(\sqrt{32}-\sqrt{50}\right)^2=25\)
\(B=1-\left(\left(-\sqrt{3}\right)^2-\left(\sqrt{20}-\sqrt{45}\right)^2\right)\)\(=1-\left(-2\right)=3\)
3) \(\sqrt{3}\left(2\sqrt{27}-\sqrt{75}+\dfrac{3}{2}\sqrt{12}\right)\)
=\(\sqrt{3}\left(2.3\sqrt{3}-5\sqrt{3}+\dfrac{3}{2}.2\sqrt{3}\right)\)
=\(\sqrt{3}\left(6\sqrt{3}-5\sqrt{3}+3\sqrt{3}\right)\)
=\(\sqrt{3}.\left(4\sqrt{3}\right)\)
=12
\(\left(2\sqrt{3}+2.3\sqrt{3}-\sqrt{3}\right):\sqrt{3}\)
=\(\left(2\sqrt{3}+6\sqrt{3}-\sqrt{3}\right):\sqrt{3}\)
=\(\left(7\sqrt{3}\right):\sqrt{3}\)
=\(7\sqrt{3}.\dfrac{1}{\sqrt{3}}\)
=7
\(1.A=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
\(2.B=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\left(3\sqrt{5}+3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\left(7-5\right)=4\) \(3.C=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\) \(4.\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\) \(5.E=\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=4+2\sqrt{3}-2\sqrt{3}+4=8\)
\(6.F=\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}=27-12\sqrt{5}+12\sqrt{5}=27\)
Lời giải:
$\sqrt{27}+\sqrt{96}-\sqrt{150}-\sqrt{12}$
$=3\sqrt{3}+4\sqrt{6}-5\sqrt{6}-2\sqrt{3}=\sqrt{3}-\sqrt{6}$
$=\sqrt{3}(1-\sqrt{2})$
Do đó: $(\sqrt{27}+\sqrt{96}-\sqrt{150}-\sqrt{12}):(1-\sqrt{2})=\sqrt{3}$