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\(=\left[\dfrac{2+\sqrt{3}}{2}:\left(1+\sqrt{\dfrac{4+2\sqrt{3}}{4}}\right)\right]+\left[\dfrac{2-\sqrt{3}}{2}:\left(1-\sqrt{\dfrac{4-2\sqrt{3}}{4}}\right)\right]\)
\(=\left(\dfrac{2+\sqrt{3}}{2}:\dfrac{2+\sqrt{3}+1}{2}\right)+\left(\dfrac{2-\sqrt{3}}{2}:\dfrac{2-\sqrt{3}+1}{2}\right)\)
\(=\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=1\)
a)\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=1\)\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=1\)
\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}=1\)
\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=1\)
\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=1\)
\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
\(\Leftrightarrow\sqrt{1}=1\) (đpcm)
a: \(VT=\dfrac{3\sqrt{6}}{2}+\dfrac{2\sqrt{6}}{3}-\dfrac{4\sqrt{6}}{2}\)
\(=\dfrac{-\sqrt{6}}{2}+\dfrac{2\sqrt{6}}{3}=\dfrac{-3\sqrt{6}+4\sqrt{6}}{6}=\dfrac{\sqrt{6}}{6}\)
b: \(VT=\dfrac{\left(\sqrt{6x}+\dfrac{\sqrt{6x}}{3}+\sqrt{6x}\right)}{\sqrt{6x}}\)
\(=1+\dfrac{1}{3}+1=2\dfrac{1}{3}\)
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+........+\frac{1}{2010\sqrt{2009}+2009\sqrt{2010}}=\frac{1}{\sqrt{1}\sqrt{2}\left(\sqrt{1}+\sqrt{2}\right)}+\frac{1}{\sqrt{2}\sqrt{3}\left(\sqrt{2}+\sqrt{3}\right)}+........+\frac{1}{\sqrt{2009}\sqrt{2010}\left(\sqrt{2009}+\sqrt{2010}\right)}\)
\(=\frac{\left(\sqrt{2010}-\sqrt{2009}\right)\left(\sqrt{2010}+\sqrt{2009}\right)}{\sqrt{2009}\sqrt{2010}\left(\sqrt{2010}+\sqrt{2009}\right)}+.......+\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\sqrt{2}\sqrt{1}\left(\sqrt{2}+\sqrt{1}\right)}=1-\frac{1}{\sqrt{2010}}=1-\frac{\sqrt{2010}}{2010}\)
Nó không bằng nhau nha bạn