题目内容
8.用归纳法化简求值:化简$\frac{1}{2\sqrt{1}+\sqrt{2}}$+$\frac{1}{3\sqrt{2}+2\sqrt{3}}$+$\frac{1}{4\sqrt{3}+3\sqrt{4}}$+…+$\frac{1}{9\sqrt{10}+10\sqrt{9}}$.分析 根据提公因式法和平方差公式得到原式=$\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2}×\sqrt{1}}$+$\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}×\sqrt{2}}$+$\frac{\sqrt{4}-\sqrt{3}}{\sqrt{4}×\sqrt{3}}$+…+$\frac{\sqrt{10}-\sqrt{9}}{\sqrt{10}×\sqrt{9}}$,再拆项抵消法求解即可.
解答 解:$\frac{1}{2\sqrt{1}+\sqrt{2}}$+$\frac{1}{3\sqrt{2}+2\sqrt{3}}$+$\frac{1}{4\sqrt{3}+3\sqrt{4}}$+…+$\frac{1}{9\sqrt{10}+10\sqrt{9}}$
=$\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2}×\sqrt{1}}$+$\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}×\sqrt{2}}$+$\frac{\sqrt{4}-\sqrt{3}}{\sqrt{4}×\sqrt{3}}$+…+$\frac{\sqrt{10}-\sqrt{9}}{\sqrt{10}×\sqrt{9}}$
=$\frac{1}{\sqrt{1}}$-$\frac{1}{\sqrt{2}}$+$\frac{1}{\sqrt{2}}$-$\frac{1}{\sqrt{3}}$+$\frac{1}{\sqrt{3}}$-$\frac{1}{\sqrt{4}}$+…+$\frac{1}{\sqrt{9}}$-$\frac{1}{\sqrt{10}}$
=$\frac{1}{\sqrt{1}}$-$\frac{1}{\sqrt{10}}$
=1-$\frac{\sqrt{10}}{10}$.
点评 此题考查了二次根式的化简求值,二次根式运算的最后,注意结果要化到最简二次根式,二次根式的乘除运算要与加减运算区分,避免互相干扰.解题的关键是将原式变形为$\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2}×\sqrt{1}}$+$\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}×\sqrt{2}}$+$\frac{\sqrt{4}-\sqrt{3}}{\sqrt{4}×\sqrt{3}}$+…+$\frac{\sqrt{10}-\sqrt{9}}{\sqrt{10}×\sqrt{9}}$.
| A. | $\frac{1}{2}$ | B. | $\frac{1}{3}$ | C. | $\frac{2}{3}$ | D. | $\frac{2}{5}$ |
| A. | (-4,0) | B. | (-2,0) | C. | (-4,0)或(-2,0) | D. | (-3,0) |
| A. | 1 | B. | 2 | C. | 3 | D. | 4 |
| A. | $\frac{\sqrt{3}}{8}$ | B. | $\frac{\sqrt{3}}{4}$ | C. | $\frac{\sqrt{2}}{4}$ | D. | $\frac{\sqrt{2}}{8}$ |
| A. | 300πcm2 | B. | 250πcm2 | C. | 200πcm2 | D. | 150πcm2 |