Giải câu 4 trang 29 sách phát triển năng lực toán 9 tập 1.
+) Ta có:
$\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}$ = $\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{[(n+1)\sqrt{n}+n\sqrt{n+1}].[(n+1)\sqrt{n}-n\sqrt{n+1}]}$ = $\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{(n+1)^{2}.n-n^{2}.(n+1)}$
= $\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{(n+1).n.(n+1-n)}$ = $\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)}=\frac{(n+1)\sqrt{n}}{n(n+1)}-\frac{n\sqrt{n+1}}{n(n+1)}$ = $\frac{\sqrt{n}}{n}-\frac{\sqrt{n+1}}{n+1}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}$
=> Điều phải chứng minh
+) S = $\frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}$
= $\frac{1}{(1+1).\sqrt{1}+1.\sqrt{1+1}}+\frac{1}{(2+1)\sqrt{2}+2\sqrt{2+1}}+\frac{1}{(3+1)\sqrt{3}+3\sqrt{3+1}}+...+\frac{1}{(99+1)\sqrt{99}+99\sqrt{99+1}}$
= $\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}$
= $1-\frac{1}{\sqrt{100}}$