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KT
6 tháng 8 2018
\(\frac{2\sqrt{12}-\sqrt{6}}{2\sqrt{6}-\sqrt{3}}+\frac{10+\sqrt{5}}{2\sqrt{15}+\sqrt{3}}\)
\(=\frac{\sqrt{2}\left(2\sqrt{6}-\sqrt{3}\right)}{2\sqrt{6}-\sqrt{3}}+\frac{\sqrt{5}\left(2\sqrt{5}+1\right)}{\sqrt{3}\left(2\sqrt{5}+1\right)}\)
\(=\sqrt{2}+\frac{\sqrt{5}}{\sqrt{3}}\)
\(=\frac{\sqrt{6}+\sqrt{5}}{\sqrt{3}}\)
p/s: chúc bạn học tốt
4 tháng 8 2018
\(\left(20.\sqrt{0.03}+12.\sqrt{3}-\frac{1}{5}.\sqrt{75}\right).\sqrt{6}\)
\(=20.\sqrt{0,03.6}+12.\sqrt{3.6}-\frac{1}{5}.\sqrt{75.6}\)
\(=20.\sqrt{\frac{9}{50}}+12.\sqrt{3^2.2}-\frac{1}{5}.\sqrt{15^2.2}\)
\(=6\sqrt{2}+36\sqrt{2}-3\sqrt{2}\)
\(=39\sqrt{2}\)
30 tháng 7 2019
\(a,\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-13\sqrt{3}\)
\(b,2\sqrt{3}.\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)
\(=2\sqrt{3}.6\sqrt{3}=36\)
\(c,\left(2\sqrt{2}-\sqrt{3}\right)^2=8-4\sqrt{6}+3\)
\(=11-4\sqrt{6}\)
\(d,\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)=1+2\sqrt{3}+3-2\)
\(=2+2\sqrt{3}\)
KT
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
7 tháng 6 2019
Thêm câu này hộ tớ nx nhé !
e) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right).\left(\sqrt{2}-3\sqrt{0.4}\right)\)
14 tháng 7 2019
\(a,\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{12}-\sqrt{6}}{2\left(\sqrt{2}-1\right)}-\frac{6\sqrt{6}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}}{2}-\frac{4\sqrt{6}}{2}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\frac{\sqrt{6}-4\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)
\(=\frac{-3\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)
\(=-\frac{3}{2}\)
5 tháng 8 2018
\(\sqrt{10-4\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{2^2-2.2.\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{3^2-2.3.2\sqrt{6}+\left(2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{6}\right)^2}+\sqrt{\left(3-2\sqrt{6}\right)^2}\)
\(=-\left(2-\sqrt{6}\right)-\left(3-2\sqrt{6}\right)\)
\(=-2+\sqrt{6}-3+2\sqrt{6}\)
\(=-5+3\sqrt{6}\)
5 tháng 8 2018
\(\sqrt{16-6\sqrt{7}}+\sqrt{32-8\sqrt{7}}\)
\(=\sqrt{3^2-2.3.\sqrt{7}+\left(\sqrt{7}\right)^2}+\sqrt{2^2-2.2.2\sqrt{7}+\left(2\sqrt{7}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{7}\right)^2}+\sqrt{\left(2-2\sqrt{7}\right)^2}\)
\(=3-\sqrt{7}-\left(2-2\sqrt{7}\right)\)
\(=3-\sqrt{7}-2+2\sqrt{7}\)
\(=1+\sqrt{7}\)
18 tháng 6 2019
\(\left(3-\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}\)
\(=\left(3-\sqrt{5}\right).\sqrt{2}\left(\sqrt{5}-1\right)\sqrt{3+\sqrt{5}}\)
\(=\left(3-\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{6+2\sqrt{5}}\)
\(=\left(3-\sqrt{5}\right)\left(\sqrt{5}-1\right)\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\left(3-\sqrt{5}\right)\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\)
\(=\left(3-\sqrt{5}\right)\left[\left(\sqrt{5}\right)^2-1\right]\)
\(=\left(3-\sqrt{5}\right)\left(5-1\right)\)
\(=4\left(3-\sqrt{5}\right)\)
\(=12-4\sqrt{5}\)
Ta có: \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
\(=\frac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\frac{\sqrt{5-2\sqrt{5}+1}+\sqrt{5+2\sqrt{5}+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{2}}=\frac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\sqrt{10}\)
Đặt \(A=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
\(\Rightarrow A^2=\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)
\(=\left(\sqrt{3-\sqrt{5}}\right)^2+2\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}+\left(\sqrt{3+\sqrt{5}}\right)^2\)
\(=\left|3-\sqrt{5}\right|+2\sqrt{9-5}+\left|3+\sqrt{5}\right|\)
\(=3-\sqrt{5}+2\sqrt{4}+3+\sqrt{5}=6+4=10\)
Vì \(\hept{\begin{cases}\sqrt{3-\sqrt{5}}>0\\\sqrt{3+\sqrt{5}}>0\end{cases}}\Rightarrow\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}>0\)
=> A > 0 mà A2 = 10 => A = √10