Ta chứng minh \(\dfrac{1}{\left(k+1\right)\sqrt{k}}+\dfrac{1}{k\sqrt{k+1}}=\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\)
Thật vậy, ta có \(\dfrac{1}{\left(k+1\right)\sqrt{k}+k\sqrt{k+1}}=\dfrac{\left(k+1\right)\sqrt{k}-k\sqrt{k+1}}{\left(k+1\right)^2k-k^2\left(k+1\right)}=\dfrac{\left(k+1\right)\sqrt{k}-k\sqrt{k+1}}{\left(k+1\right)k}=\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\)(luôn đúng)
Khi đó, ta có \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2010\sqrt{2009}+2009\sqrt{2010}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2009}}-\dfrac{1}{\sqrt{2010}}=1-\dfrac{1}{\sqrt{2010}}=\dfrac{\sqrt{2010}-1}{\sqrt{2010}}\)
