Rút gọn
\(\frac{\sqrt{160}-\sqrt{80}}{\sqrt{8}-\sqrt{2}}-\frac{\sqrt{40}-\sqrt{15}}{2\sqrt{2}-\sqrt{3}}\)
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BA
24 tháng 11 2019
\(a,A=\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}\)
\(=\sqrt{\left(\sqrt{5}^2+2\sqrt{5}+2\sqrt{2}\cdot\sqrt{5}\right)+\sqrt{2}^2+2\sqrt{2}\cdot1+1^2}\)
\(=\sqrt{\sqrt{5}^2+2\cdot\sqrt{5}\left(\sqrt{2}+1\right)+\left(\sqrt{2}+1\right)^2}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{2}+1\right)^2}\)
\(=\sqrt{5}+\sqrt{2}+1\)
\(b,B=\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\frac{3\cdot\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}{\sqrt{6}+1}+\frac{2\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}{\sqrt{6}-2}-\frac{4\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left[3\cdot\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}+11\right)\left(\sqrt{6}-11\right)=-115\)
24 tháng 7 2016
2) \(A=\sqrt{15a^2-8a\sqrt{15}+16}\\ =\sqrt{\left(a\sqrt{15}-4\right)^2}\)
b) Khi a=\(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\) thì
\(A=\sqrt{\left[\left(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\right)\sqrt{15}-4\right]^2}\)
\(=\sqrt{\left[\left(3+5\right)-4\right]^2}\)
\(=\sqrt{4^2}\)
\(=4\)
CD
1
CH
Cô Hoàng Huyền
Admin
VIP
14 tháng 6 2017
a) \(A=\frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{2\sqrt{2}+\sqrt{6}}{3+\sqrt{3}}+\frac{2\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}\)
\(=\frac{6\sqrt{2}-2\sqrt{6}+3\sqrt{6}-\sqrt{18}+6\sqrt{2}+2\sqrt{6}-3\sqrt{6}-\sqrt{18}}{6}\)
\(=\frac{12\sqrt{2}-2\sqrt{18}}{6}=\frac{6\sqrt{2}}{6}=\sqrt{2}\)
27 tháng 8 2020
1) \(A^2=2+2.\frac{\sqrt{\left(8+\sqrt{15}\right)\left(8-\sqrt{15}\right)}}{2}\)
\(2+\sqrt{64-15}=2+\sqrt{49}=2+7=9\) mà A>0
=> A=3
28 tháng 8 2020
2) \(A=\sqrt{4-\sqrt{15}}\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right).\)
\(A=\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right).\)
\(A=\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right).\)
\(A^2=\left(4+\sqrt{15}\right)\left(16-4\sqrt{15}\right)\)
\(=4\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=4\)
Mà A >0
=> A=2
Mà 4>3
=> \(\sqrt{4}=2>\sqrt{3}\)
=> \(A>\sqrt{3}\)
12 tháng 8 2018
\(A=\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)\)
\(A=\left(\frac{5+2\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}+\frac{5-2\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\left(15+2\sqrt{6}\right)\)
\(A=\left(\frac{5+2\sqrt{6}+5-2\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\left(15+2\sqrt{6}\right)\)
\(A=\left(\frac{10}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\left(15+2\sqrt{6}\right)\)
\(A=\left(\frac{10}{25-24}\right)\left(15+2\sqrt{6}\right)\)
\(A=10\left(15+2\sqrt{6}\right)\)
\(A=150+20\sqrt{6}\)
1 tháng 10 2016
\(D=\left(\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}-\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\right).\frac{\sqrt{3}-1}{3-\sqrt{3}}=\left(\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{-1}-\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{-1}\right).\frac{\sqrt{3}-1}{\sqrt{3}\left(\sqrt{3}-1\right)}\)
\(=\left(-5-2\sqrt{6}+5-2\sqrt{6}\right).\frac{1}{\sqrt{3}}=\frac{-4\sqrt{6}}{\sqrt{3}}=-4\sqrt{2}\)
\(E=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}=\frac{\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-2\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\frac{\sqrt{3}+\sqrt{5}+\sqrt{5}-\sqrt{3}-2\sqrt{5}+2}{\sqrt{2}}\)
\(=\sqrt{2}\)
12 tháng 5 2018
b) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right).\sqrt{2}.\left(\sqrt{5}-\sqrt{3}\right).\sqrt{4-\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}\)
\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)\)
\(=\left(4+\sqrt{15}\right).2\)
\(=8+2\sqrt{15}\)
12 tháng 5 2018
a) \(\frac{3\sqrt{8}-2\sqrt{12}+\sqrt{40}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)
\(=\frac{6\sqrt{2}-4\sqrt{3}+2\sqrt{10}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)
\(=\frac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{10}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}\)
\(=\frac{2\sqrt{2}}{3}\)
