\(\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\)
VP = \(\frac{1}{\sqrt{n+1}+\sqrt{n}}\)
= \(\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)
= \(\frac{\sqrt{n+1}-\sqrt{n}}{\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}\right)^2}\)
= \(\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}\)
= \(\sqrt{n+1}-\sqrt{n}\)
= VT
Vậy đẳng thức được chứng minh
cách khác nhé:
Xét: \(\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)\)
\(=\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}\right)^2\)
\(=n+1-n=1\)
\(\Rightarrow\)\(\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\) (đpcm)
