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a) \(\left(2\sqrt{2}-3\right)^2\)
\(=\left(2\sqrt{2}\right)^2-2\cdot2\sqrt{2}\cdot3+3^2\)
\(=4\cdot2-12\sqrt{2}+9\)
\(=17-12\sqrt{2}\)
b) \(\sqrt{\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\right)^2}\)
\(=\left|\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\right|\)
\(=\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\)
\(=\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\)
\(=\dfrac{\sqrt{2}-1}{2}\)
c) \(\sqrt{\left(0,1-\sqrt{0,1}\right)^2}\)
\(=\left|0,1-\sqrt{0,1}\right|\)
\(=0,1-\sqrt{0,1}\)
a: \(\dfrac{5+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}=\dfrac{\left(5+2\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)}{3}=\dfrac{5\sqrt{5}-5\sqrt{2}+10-2\sqrt{10}}{3}\)
b: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\)
a,Ta có : \(1-\sqrt{3}\); \(\sqrt{2}-\sqrt{6}=\sqrt{2}\left(1-\sqrt{3}\right)\Rightarrow1-\sqrt{3}< \sqrt{2}\left(1-\sqrt{3}\right)\)
Vậy \(1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)
b, Đặt A = \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)(*)
\(\sqrt{2}A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)
\(=\sqrt{7}+1-\sqrt{7}+1-2=0\Rightarrow A=0\)
Vậy (*) = 0
1:
Ta có: \(\sqrt{2}-\sqrt{6}\)
\(=\sqrt{2}\left(1-\sqrt{3}\right)< 0\)
\(\Leftrightarrow1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)
a) \(\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{2}\right|+\left|\sqrt{5}+\sqrt{2}\right|\)
\(=\sqrt{5}-\sqrt{2}+\sqrt{5}+\sqrt{2}\)
\(=\sqrt{5}+\sqrt{5}\)
\(=2\sqrt{5}\)
b) \(\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(\sqrt{2}-5\right)^2}\)
\(=\left|\sqrt{2}-1\right|-\left|\sqrt{2}-5\right|\)
\(=\sqrt{2}-1-\left(5-\sqrt{2}\right)\)
\(=\sqrt{2}-1-5+\sqrt{2}\)
\(=2\sqrt{2}-6\)