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HT
2
29 tháng 7 2015
a/ \(\left(\sqrt{6}+2\right)\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{2}\left(\sqrt{3}+2\right)\left(\sqrt{3.1}-\sqrt{2.1}\right)=\sqrt{2}\left(3-2\right)=\sqrt{2}\)
b/ \(\left(\sqrt{\frac{8}{3}-\sqrt{24}+\sqrt{\frac{50}{3}}}\right)=-\frac{4\sqrt{6}}{3}+\frac{5\sqrt{6}}{3}=\frac{\sqrt{6}}{3}\)
PA
25 tháng 6 2017
b)
\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)=6-121=-115\)
PA
25 tháng 6 2017
a)
\(\dfrac{\left(3\sqrt{3}+5\sqrt{2}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
\(=\dfrac{\left(3\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)
\(=\dfrac{\left(3\sqrt{3}+5\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{5\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\dfrac{9-3\sqrt{6}+5\sqrt{6}-10}{5}=\dfrac{-1+2\sqrt{6}}{5}\)
PT
3
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,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)
\(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)
19 tháng 7 2018
\(C=\sqrt{15-6\sqrt{6}}+\sqrt{33+12\sqrt{6}}=\sqrt{9-2.3\sqrt{6}+6}+\sqrt{24+2.3.2\sqrt{6}+9}=3-\sqrt{6}+2\sqrt{6}+3=6+\sqrt{6}\) \(D=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{3-2\sqrt{3}+1}-\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\) \(F=\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\)
19 tháng 7 2018
\(C=\sqrt{15-6\sqrt{6}}+\sqrt{33+12\sqrt{6}}=\sqrt{\left(\sqrt{9}-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}+\sqrt{9}\right)^2}=3-\sqrt{6}+2\sqrt{6}+3=6+\sqrt{6}\)
\(D=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(\Rightarrow\sqrt{2}D=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}-1-\sqrt{3}-1=-2\)
\(\Rightarrow D=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\)
\(F=\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=\left(3\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2=27-2=25\)
25 tháng 11 2019
a. \(\sqrt{50}-\sqrt{3}.\sqrt{6}+\frac{\sqrt{22}}{\sqrt{11}}=5\sqrt{2}-3\sqrt{2}+\sqrt{2}=3\sqrt{2}\)
b. \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\sqrt{7+4\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}=\sqrt{2}\)
22 tháng 9 2021
\(6\sqrt{\dfrac{1}{2}}+\dfrac{2}{\sqrt{2}}-\sqrt{50}=3\sqrt{2}+\sqrt{2}-5\sqrt{2}=-\sqrt{2}\\ \dfrac{2+\sqrt{3}}{2-\sqrt{3}}-\dfrac{2-\sqrt{3}}{2+\sqrt{3}}=\dfrac{\left(2+\sqrt{3}\right)^2-\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\\ =\dfrac{7+4\sqrt{3}-\left(7-4\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\dfrac{8\sqrt{3}}{1}=8\sqrt{3}\)
(2√3 - √50) . √3 + 3√6
= 18 - 5√6 + 3√6
= 18 - 2√6