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6 tháng 8 2018

\(\left(\frac{3+2\sqrt{3}}{\sqrt{3}+2}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\right):\left(\sqrt{2}+\sqrt{3}\right)\)

\(=\left(\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}+2}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\right):\left(\sqrt{2}+\sqrt{3}\right)\)

\(=\left(\sqrt{3}+\sqrt{2}\right):\left(\sqrt{2}+\sqrt{3}\right)\)

\(=1\)

p/s: chúc bạn học tốt

26 tháng 10 2020

a) \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{1+\sqrt{2}}-2+\sqrt{3}\)

\(=\frac{\sqrt{3}.\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}.\left(\sqrt{2}+1\right)}{1+\sqrt{2}}-2+\sqrt{3}\)

\(=\sqrt{3}+2+\sqrt{2}-2+\sqrt{3}\)

\(=2\sqrt{3}+\sqrt{2}\)

b) \(\frac{-3}{2}.\sqrt{9-4\sqrt{5}}+\sqrt{\left(-4\right)^2.\left(1+\sqrt{5}\right)^2}\)

\(=\frac{-3}{2}.\sqrt{5-4\sqrt{5}+4}+\sqrt{4^2.\left(1+\sqrt{5}\right)^2}\)

\(=\frac{-3}{2}.\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{4^2}.\sqrt{\left(1+\sqrt{5}\right)^2}\)

\(=\frac{-3}{2}.\left|\sqrt{5}-2\right|+4.\left|1+\sqrt{5}\right|\)

\(=\frac{-3}{2}.\left(\sqrt{5}-2\right)+4\left(1+\sqrt{5}\right)\)

\(=\frac{-3\sqrt{5}}{2}+3+4+4\sqrt{5}\)

\(=\frac{-3\sqrt{5}}{2}+4\sqrt{5}+7\)

\(=\frac{-3\sqrt{5}}{2}+\frac{8\sqrt{5}}{2}+\frac{14}{2}\)

\(=\frac{-3\sqrt{5}+8\sqrt{5}+14}{2}=\frac{14+5\sqrt{5}}{2}\)

12 tháng 10 2020

a) \(\left(\frac{\sqrt{9}}{2}+\frac{\sqrt{1}}{2}-\sqrt{2}\right)\sqrt{2}\)

\(=\frac{3\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-2\)

\(=\frac{4\sqrt{2}}{2}-2=2\sqrt{2}-2\)

b) \(\left(\frac{\sqrt{8}}{3}-\sqrt{24}+\frac{\sqrt{50}}{3}\right)\sqrt{6}\)

\(=\frac{4\sqrt{3}}{3}-12+\frac{10\sqrt{3}}{3}\)

\(=\frac{14\sqrt{3}}{3}-12\)

c) \(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{1}\right)\) (đã sửa đề)

\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\sqrt{2}\)

\(=\left(3-1\right)\sqrt{2}\)

\(=2\sqrt{2}\)

d) \(\left(3\sqrt{2}+1\right)\left(\sqrt{3\sqrt{2}-1}\right)\)

\(=\sqrt{3\sqrt{2}+1}\cdot\left(\sqrt{3\sqrt{2}+1}\cdot\sqrt{3\sqrt{2}-1}\right)\)

\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{18-1}\)

\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{17}\)

...

28 tháng 7 2016

\(\left(\sqrt{4,5}-\frac{1}{2}.\sqrt{72}+5\sqrt{\frac{1}{2}}\right).\left(42\sqrt{\frac{25}{6}}-10\sqrt{\frac{3}{2}}-12\sqrt{\frac{98}{3}}\right)\)

=\(\left(\frac{3\sqrt{2}}{2}-3\sqrt{2}+\frac{5\sqrt{2}}{2}\right).\left(35\sqrt{6}-5\sqrt{6}-28\sqrt{6}\right)\)

=\(\left(\frac{3\sqrt{2}-6\sqrt{2}+5\sqrt{2}}{2}\right).2\sqrt{6}\)

=\(2\sqrt{2}.\sqrt{6}=4\sqrt{3}\)

20 tháng 8 2020

Ta có:

\(B=\frac{\sqrt{2+\sqrt{3}}}{2}\div\left(\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)

\(B=\frac{\sqrt{4+2\sqrt{3}}}{2}\div\left(\frac{\sqrt{4+2\sqrt{3}}}{2}-\frac{2\sqrt{3}}{3}+\frac{\sqrt{4+2\sqrt{3}}}{2\sqrt{3}}\right)\)

\(B=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}\div\left(\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}-\frac{2\sqrt{3}}{3}+\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2\sqrt{3}}\right)\)

\(B=\frac{\sqrt{3}+1}{2}\div\left(\frac{\sqrt{3}+1}{2}-\frac{2\sqrt{3}}{2}+\frac{\left(\sqrt{3}+1\right)\sqrt{3}}{6}\right)\)

\(B=\frac{\sqrt{3}+1}{2}\div\left[\frac{3\left(\sqrt{3}+1\right)-6\sqrt{3}+3+\sqrt{3}}{6}\right]\)

\(B=\frac{\sqrt{3}+1}{2}\div\frac{6-2\sqrt{3}}{6}\)

\(B=\frac{\sqrt{3}+1}{2}.\frac{6}{6-2\sqrt{3}}\)

\(B=\frac{3+2\sqrt{3}}{2}\)