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\(a,\sqrt{\frac{72}{9}}:\sqrt{8}=\frac{\sqrt{72}}{\sqrt{9}}.\frac{1}{\sqrt{8}}\)
\(=\frac{6\sqrt{2}}{3}.\frac{1}{2\sqrt{2}}\)
\(=1\)
\(b,\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right):\sqrt{3}=\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right):\sqrt{3}\)
\(=33\sqrt{3}:\sqrt{3}\)
\(=33\)
\(c,\left(\sqrt{125}+\sqrt{245}-\sqrt{5}\right):\sqrt{5}=\left(5\sqrt{5}+7\sqrt{5}-\sqrt{5}\right):\sqrt{5}\)
\(=11\sqrt{5}:\sqrt{5}\)
\(=11\)
\(d,\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{7}\right):\sqrt{7}=\left(\frac{1}{\sqrt{7}}-\frac{4}{\sqrt{7}}+\frac{7}{\sqrt{7}}\right):\sqrt{7}\)
\(=\frac{4}{\sqrt{7}}.\frac{1}{\sqrt{7}}=\frac{4}{7}\)
a/ Đề sai
b/ \(\sqrt{125}-4\sqrt{45}+3\sqrt{2}-\sqrt{80}=5\sqrt{5}-12\sqrt{5}+3\sqrt{2}-4\sqrt{5}\)
\(=-11\sqrt{5}+3\sqrt{2}\)
c/ \(2\sqrt{\frac{27}{4}}-\sqrt{\frac{48}{9}}-\frac{2}{5}\sqrt{\frac{75}{16}}=2.\frac{3\sqrt{3}}{2}-\frac{4\sqrt{3}}{3}-\frac{2}{5}.\frac{5\sqrt{3}}{4}\)
\(=3\sqrt{3}-\frac{4\sqrt{3}}{3}-\frac{\sqrt{3}}{2}=\sqrt{3}\left(3-\frac{4}{3}-\frac{1}{2}\right)=\frac{7\sqrt{3}}{6}\)
d/ \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\cdot\sqrt{11}+3\sqrt{22}=33-3\sqrt{22}-11+3\sqrt{22}=22\)
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)\(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)\)
\(=2\sqrt{3}\left(\sqrt{9\cdot3}+2\sqrt{16\cdot3}-\sqrt{25\cdot3}\right)\)
\(=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)
\(2\sqrt{3}\cdot6\sqrt{3}=12\cdot3=36\)
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/ \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{13-4\sqrt{10}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{8-4\sqrt{10}+5}-\sqrt{45+12\sqrt{10}+8}\)
\(=\sqrt{\left(2\sqrt{2}\right)^2-2\cdot2\sqrt{2\cdot5}+\left(\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\cdot2\sqrt{5\cdot2}+\left(2\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)
\(=2\sqrt{2}-\sqrt{5}-3\sqrt{5}-2\sqrt{2}\)
\(=-4\sqrt{5}\)
\(a,\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-13\sqrt{3}\)
\(b,2\sqrt{3}.\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)
\(=2\sqrt{3}.6\sqrt{3}=36\)
\(c,\left(2\sqrt{2}-\sqrt{3}\right)^2=8-4\sqrt{6}+3\)
\(=11-4\sqrt{6}\)
\(d,\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)=1+2\sqrt{3}+3-2\)
\(=2+2\sqrt{3}\)
1 , \(\left(\sqrt{12}-2\sqrt{75}\right).\sqrt{3}=\sqrt{12.3}-\sqrt{300.3}=6-30=-24\)
2 , \(\sqrt{3}.\left(\sqrt{12}.\sqrt{27}-\sqrt{3}\right)=\sqrt{12.27.3}-\sqrt{3.3}=18\sqrt{3}-3\)
3 , \(\left(7\sqrt{48}+3\sqrt{27}-\sqrt{12}\right):\sqrt{3}=\left(28\sqrt{3}+9\sqrt{3}-2\sqrt{3}\right):\sqrt{3}=35\)
4 , bạn làm tương tự nhé
5 , bạn làm tương tự nhé
6 , bạn làm tương tự nhé
= ( 7\(\sqrt{16\cdot3}\)+3\(\sqrt{9\cdot3}\)-3\(\sqrt{4\cdot3}\)) /\(\sqrt{3}\)
=(49\(\sqrt{3}\)+ 9\(\sqrt{3}\)-6 \(\sqrt{3}\)) /\(\sqrt{3}\)
=52