Tính :
a) (\(\sqrt{12}+\sqrt{75}+\sqrt{27}\)) / \(\sqrt{15}\)
b) \(\frac{\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right)}{\sqrt{10}}\)
c)\(\frac{\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{\frac{9}{7}}\right)}{7}\)
K
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14 tháng 7 2020
Bài 1:
a) Ta có: \(\frac{12\sqrt{50}-8\sqrt{200}+7\sqrt{450}}{\sqrt{10}}\)
\(=\frac{12\cdot\sqrt{5}\cdot\sqrt{10}-8\cdot\sqrt{20}\cdot\sqrt{10}+7\cdot\sqrt{45}\cdot\sqrt{10}}{\sqrt{10}}\)
\(=\frac{\sqrt{10}\left(12\sqrt{5}-8\sqrt{20}+7\sqrt{45}\right)}{\sqrt{10}}\)
\(=12\sqrt{5}-8\sqrt{20}+7\sqrt{45}\)
\(=\sqrt{5}\left(12-16+21\right)\)
\(=17\sqrt{5}\)
b) Ta có: \(\frac{\frac{\sqrt{1}}{7}-\sqrt{\frac{16}{7}}+\sqrt{\frac{9}{7}}}{\sqrt{7}}\)
\(=\left(\frac{1}{\sqrt{7}}-\frac{4}{\sqrt{7}}+\frac{3}{\sqrt{7}}\right)\cdot\frac{1}{\sqrt{7}}\)
\(=0\cdot\frac{1}{\sqrt{7}}=0\)
2 tháng 6 2017
- \(\frac{\sqrt{27\left(1-\sqrt{3}\right)^4}}{3\sqrt{15}}=\frac{\sqrt{3.3^2\left(1-\sqrt{3}\right)^4}}{3\sqrt{15}}=\frac{3\left(1-\sqrt{3}\right)^2}{3\sqrt{15}}=\frac{1-2\sqrt{3}+3}{\sqrt{15}}=\frac{4-2\sqrt{3}}{\sqrt{15}}\)
- \(=\frac{\sqrt{10}\left(12-8\sqrt{2}+7.15\sqrt{2}\right)}{\sqrt{10}}=12+97\sqrt{2}\)
- \(=\sqrt{\frac{x.x\sqrt{y}}{y}}=\sqrt{\frac{x^2}{\sqrt{y}}}=\frac{|x|}{\sqrt[4]{y}}\)
NC
18 tháng 7 2018
a) \(\sqrt{200}+2\sqrt{108}-\sqrt{98}+\frac{1}{3}\sqrt{\frac{81}{3}}-3\sqrt{75}\)
\(=10\sqrt{2}+12\sqrt{3}-7\sqrt{2}+\sqrt{3}-15\sqrt{3}\)
\(=3\sqrt{2}-2\sqrt{3}\)
b)\(\left(21\sqrt{\frac{1}{7}}+\frac{1}{2}\sqrt{112}-\frac{14}{3}\sqrt{\frac{9}{7}}+7\right):3\sqrt{7}\)
\(=\left(3\sqrt{7}+2\sqrt{7}-2\sqrt{7}+7\right):3\sqrt{7}\)
\(=\frac{\sqrt{7}\left(3+\sqrt{7}\right)}{3\sqrt{7}}=\frac{\sqrt{7}+3}{3}\)
c)\(\left(\sqrt{27}-\sqrt{125}+\sqrt{45}+\sqrt{12}\right)\left(\sqrt{75}+\sqrt{20}\right)\)
\(=\left(3\sqrt{3}-5\sqrt{5}+3\sqrt{5}+2\sqrt{3}\right)\left(5\sqrt{3}+2\sqrt{5}\right)\)
\(=\left(5\sqrt{3}-2\sqrt{5}\right)\left(5\sqrt{3}+2\sqrt{5}\right)\)
\(=75-20=55\)
d)\(\left(\frac{3}{\sqrt{6}-3}-\frac{3}{\sqrt{6}+3}\right).\frac{3-\sqrt{3}}{2-2\sqrt{3}}-\frac{\sqrt{28-6\sqrt{3}}}{1}\)
\(=\frac{3\left(\sqrt{6}+3\right)-3\left(\sqrt{6}-3\right)}{-3}.\frac{3-\sqrt{3}}{2-2\sqrt{3}}-\sqrt{\left(3\sqrt{3}-1\right)^2}\)
\(=\frac{-6\left(3-\sqrt{3}\right)}{2-2\sqrt{3}}-\left(3\sqrt{3}-1\right)\left(do3\sqrt{3}>1\right)\)
\(=\frac{6\sqrt{3}-18}{2-2\sqrt{3}}-\frac{8\sqrt{3}-20}{2-2\sqrt{3}}\)
\(=\frac{6\sqrt{3}-18-8\sqrt{3}+20}{2-2\sqrt{3}}=\frac{2-2\sqrt{3}}{2-2\sqrt{3}}=1\)
2 tháng 7 2021
a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)
\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)
\(=33\sqrt{3}\cdot\sqrt{3}\)
=99
b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)
\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)
c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=36-36\sqrt{2}+18\sqrt{3}\)
d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)
\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)
2 tháng 7 2021
a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)
\(=28.3+9.3-4.3=99\)
b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)
\(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)
11 tháng 8 2021
a: Ta có: \(\sqrt{75}-2\sqrt{27}+\sqrt{48}\)
\(=5\sqrt{3}-2\cdot3\sqrt{3}+4\sqrt{3}\)
\(=3\sqrt{3}\)
c: Ta có: \(\sqrt{8+2\sqrt{7}}-\sqrt{11-4\sqrt{7}}\)
\(=\sqrt{7}+1-\sqrt{7}+2\)
=3
a)\(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}\)
\(=\left(2\sqrt{3}+5\sqrt{3}+3\sqrt{3}\right):\sqrt{15}\)
\(=10\sqrt{3}:\sqrt{15}=\sqrt{300}:\sqrt{15}=\sqrt{20}=2\sqrt{5}\)
b) \(\frac{12\sqrt{50}-8\sqrt{200}+7\sqrt{450}}{\sqrt{10}}\)
\(=\frac{60\sqrt{2}-80\sqrt{2}+105\sqrt{2}}{\sqrt{10}}\)
\(=\frac{85\sqrt{2}}{10}=\frac{17\sqrt{2}}{2}\)
c)\(\frac{\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{\frac{9}{7}}}{7}=\frac{\frac{1}{\sqrt{7}}-\frac{4}{\sqrt{7}}+\frac{3}{\sqrt{7}}}{7}=\frac{0}{7}=0\)