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1: \(=\sqrt{5}-2-3-\sqrt{5}=-5\)
2: \(=3\sqrt{2}+\sqrt{10}+3\sqrt{2}-\sqrt{10}=6\sqrt{2}\)
3: \(=2\sqrt{2}-\sqrt{5}-4\sqrt{3}-\sqrt{5}=2\sqrt{2}-4\sqrt{3}-2\sqrt{5}\)
1: \(=\sqrt{5}-2-3-\sqrt{5}=-5\)
2: \(=3\sqrt{2}+\sqrt{10}+3\sqrt{2}-\sqrt{10}=6\sqrt{2}\)
3: \(=2\sqrt{2}-\sqrt{5}-4\sqrt{3}-\sqrt{5}=2\sqrt{2}-4\sqrt{3}-2\sqrt{5}\)
81 ; 11660 ; 2 ; 4 ; 6 ; 8 ; 10 ; 12 ; 14 ; 16 ; 18 ; 20 ; 34 ; 42 .
Bài 1:
a) Ta có: \(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{45-2\cdot\sqrt{45}\cdot1+1}-\sqrt{9-2\cdot\sqrt{9}\cdot\sqrt{20}+20}\)
\(=\sqrt{\left(\sqrt{45}-1\right)^2}-\sqrt{\left(3-\sqrt{20}\right)^2}\)
\(=\left|\sqrt{45}-1\right|-\left|3-\sqrt{20}\right|\)
\(=\sqrt{45}-1-3+\sqrt{20}\)
\(=\sqrt{45}+\sqrt{20}-4\)
\(=\sqrt{5}\left(3+2\right)-4=5\sqrt{5}-4\)
b) Ta có: \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{8}+8}-\sqrt{45+2\cdot\sqrt{45}\cdot\sqrt{8}+8}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{8}\right)^2}-\sqrt{\left(\sqrt{45}+\sqrt{8}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{8}\right|-\left|\sqrt{45}+\sqrt{8}\right|\)
\(=\sqrt{8}-\sqrt{5}-\sqrt{45}-\sqrt{8}\)
\(=-\sqrt{5}-\sqrt{45}=-\sqrt{5}\left(1+\sqrt{9}\right)=-4\sqrt{5}\)
c) Ta có: \(\left(3-\sqrt{2}\right)\cdot\sqrt{7+4\sqrt{3}}\)
\(=\left(3-\sqrt{2}\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot2+4}\)
\(=\left(3-\sqrt{2}\right)\cdot\sqrt{\left(\sqrt{3}+2\right)^2}\)
\(=\left(3-\sqrt{2}\right)\left(\sqrt{3}+2\right)\)
\(=3\sqrt{3}+6-\sqrt{6}-2\sqrt{2}\)
d) Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10+2\sqrt{21}}\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\cdot\sqrt{7+2\cdot\sqrt{7}\cdot\sqrt{3}+3}\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\cdot\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\cdot\left(\sqrt{7}+\sqrt{3}\right)\)
\(=\left(\sqrt{7}\right)^2-\left(\sqrt{3}\right)^2=7-3=4\)
\(A=\left(2-\sqrt{3}\right)\sqrt{4+2.2.\sqrt{3}+3}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=1\)
các câu còn lại làm tương tự nhé bạn !
5) Ta có: \(\dfrac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)
\(=\dfrac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{5\left(\sqrt{3}-\sqrt{2}\right)}\)
=1
1. \(\sqrt{\left(9-4\sqrt{5}\right)}\) - \(\sqrt{\left(14+6\sqrt{5}\right)}\) = \(\sqrt{5+4-2\cdot2\sqrt{5}}\) - \(\sqrt{9+5+2\cdot3\sqrt{5}}\) = \(\sqrt{\left(2-\sqrt{5}\right)^2}\) - \(\sqrt{\left(3+\sqrt{5}\right)^2}\) = \(\sqrt{5}-2\) - \(3-\sqrt{5}\) = -5
1: \(=\sqrt{5}-2-3-\sqrt{5}=-5\)
2: \(=3\sqrt{2}+\sqrt{10}+3\sqrt{2}-\sqrt{10}=6\sqrt{2}\)
4: \(=2\sqrt{2}-\sqrt{5}-4\sqrt{3}-\sqrt{5}=2\sqrt{2}-4\sqrt{3}-2\sqrt{5}\)