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\(A=\sqrt{12}+2\sqrt{27}+3\sqrt{45}-9\sqrt{48}\)
\(=2\sqrt{3}+6\sqrt{3}+9\sqrt{5}-36\sqrt{3}\)
\(=9\sqrt{5}-28\sqrt{3}\)
\(B=\left(\sqrt{48}-2\sqrt{75}+\sqrt{108}-\sqrt{147}\right):\sqrt{3}\)
\(=4-2\cdot5+6-7\)
\(=4-10+6-7\)
=-7
A=\(\sqrt{12}\)+2\(\sqrt{27}\)+3\(\sqrt{45}\) -9\(\sqrt{48}\)
=\(\sqrt{4.3}\) +2\(\sqrt{9.3}\)+3\(\sqrt{9.5}\) -9\(\sqrt{16.3}\)
=2\(\sqrt{3}\) +6\(\sqrt{3}\)+9\(\sqrt{5}\) -36\(\sqrt{3}\)
=\(\sqrt{3}\)(2+6-36) + 9\(\sqrt{5}\)
=9\(\sqrt{5}\)- 28\(\sqrt{3}\)
a/\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)
b/ \(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)\)
\(=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)
\(=2\sqrt{3}\cdot6\sqrt{3}=2\cdot6\cdot3=36\)
c/ \(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)\)
\(=\left(1+\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2\)
\(=1+2\sqrt{3}+3-2\)
\(=2+2\sqrt{3}\)
d/ \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{13-4\sqrt{10}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{8-4\sqrt{10}+5}-\sqrt{45+12\sqrt{10}+8}\)
\(=\sqrt{\left(2\sqrt{2}\right)^2-2\cdot2\sqrt{2\cdot5}+\left(\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\cdot2\sqrt{5\cdot2}+\left(2\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)
\(=2\sqrt{2}-\sqrt{5}-3\sqrt{5}-2\sqrt{2}\)
\(=-4\sqrt{5}\)
a) \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=\left(2+6+15-36\right)\sqrt{3}=-13\sqrt{3}\)
b) \(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)=6\left(3+8-5\right)=36\)
a)\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
\(=\sqrt{4\cdot3}+2\sqrt{9\cdot3}+3\sqrt{25\cdot3}-9\sqrt{16\cdot3}\)
\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-13\sqrt{3}\)
b)\(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)\)
\(=2\sqrt{3}\left(\sqrt{9\cdot3}+2\sqrt{16\cdot3}-\sqrt{25\cdot3}\right)\)
\(=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)
\(2\sqrt{3}\cdot6\sqrt{3}=12\cdot3=36\)
Ta có: \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-13\sqrt{3}\)
\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\\ =2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}=-13\sqrt{3}\)
a: \(A=6\sqrt{3}+10\sqrt{3}-12\sqrt{3}=4\sqrt{3}\)
b: \(B=7\sqrt{3}+5\sqrt{3}-12\sqrt{3}=0\)
c: \(=12\sqrt{2}-6+3\left(9-4\sqrt{2}\right)=12\sqrt{2}-6+27-12\sqrt{2}=21\)
d: \(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=4\sqrt{5}\)
\(=2\sqrt{80\sqrt{3}}+3\sqrt{20\sqrt{3}}-2\sqrt{5\sqrt{3}}-4\sqrt{45\sqrt{3}}\)
\(=8\sqrt{5\sqrt{3}}+6\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-12\sqrt{5\sqrt{3}}\)
=0
a) \(E=2\sqrt{40\sqrt{12}}+3\sqrt{5\sqrt{48}}-2\sqrt{\sqrt{75}}-4\sqrt{15\sqrt{27}}.\)
\(=8\sqrt{5\sqrt{3}}+6\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}-12\sqrt{5\sqrt{3}}}\)
\(=0\)
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 \(F=\frac{2\sqrt{3}+\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}\)
a: \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-14\sqrt{3}\)
b: \(\left(2\sqrt{2}-\sqrt{3}\right)^2=11-4\sqrt{6}\)