C/m các đẳng thức sau:
a) \(\sqrt{21-6\sqrt{6}}\) + \(\sqrt{9+2\sqrt{18}}\) - \(2\sqrt{6+3\sqrt{3}}\) = 0
b) \(\dfrac{1}{\sqrt{25}+\sqrt{24}}\) + \(\dfrac{1}{\sqrt{24}+\sqrt{23}}\) \(\dfrac{1}{\sqrt{23}+\sqrt{22}}\) +...+ \(\dfrac{1}{\sqrt{2}+\sqrt{1}}\) = 4
c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\) = \(\sqrt{2}\) - 1
Mn giúp mk với !!!
Câu b nhé:
Ta có:
\(\dfrac{1}{\sqrt{25}+\sqrt{24}}+\dfrac{1}{\sqrt{24}+\sqrt{23}}+\dfrac{1}{\sqrt{23}+\sqrt{22}}+...+\dfrac{1}{\sqrt{2}+\sqrt{1}}\\ =\dfrac{\sqrt{25}-\sqrt{24}}{\left(\sqrt{25}+\sqrt{24}\right)\left(\sqrt{25}-\sqrt{24}\right)}+\dfrac{\sqrt{24}-\sqrt{23}}{\left(\sqrt{24}+\sqrt{23}\right)\left(\sqrt{24}-\sqrt{23}\right)}+...+\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}+\sqrt{1}\right)\left(\sqrt{2}-\sqrt{1}\right)}\\ =\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-\sqrt{1}\\ =5-1=4\left(đpcm\right)\)
a) \(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}=0\) (*)
\(\Leftrightarrow\left(3\sqrt{2}-\sqrt{3}\right)+\left(\sqrt{3}+\sqrt{6}\right)-\left(3+\sqrt{3}\right)\cdot\sqrt{2}=0\)
\(\Leftrightarrow0=0\) (luôn đúng)
Vậy (*) luôn đúng