Cm : 1/căn1 + 1/căn2 + 1/căn3 +...+ 1/căn32 > 10
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\(\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}-\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}}{3-2}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}-\sqrt{3}\)
\(=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}-\sqrt{3}\)
\(=2\sqrt{3}-\sqrt{3}\)
\(=\sqrt{3}\)
\(\sqrt{3}-\frac{5}{2}>\sqrt{3}-4\text{ vì }-\frac{5}{2}>-4\)
\(\Rightarrow2.\left(\sqrt{3}-\frac{5}{2}\right)>\sqrt{3}-4\)
\(\Rightarrow2.\sqrt{3}-5>\sqrt{3}-4\)
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: \(\sqrt{2}-\sqrt{6}\)
\(=\sqrt{2}\left(1-\sqrt{3}\right)< 0\)
\(\Leftrightarrow1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)