问题266: 计算 1/√3+2√2 +1/√5+2√6+1/√7+4√3+...+1/√99+70√2
计算
$$ \frac{1}{\sqrt{3+2 \sqrt{2}}}+\frac{1}{\sqrt{5+2 \sqrt{6}}}+\frac{1}{\sqrt{7+4 \sqrt{3}}} \, \cdots+\frac{1}{\sqrt{99+70 \sqrt{2}}} $$
$$ \begin{aligned} & \frac{1}{\sqrt{3+2 \sqrt{2}}}+\frac{1}{\sqrt{5+2 \sqrt{6}}}+\frac{1}{\sqrt{7+4 \sqrt{3}}}+\cdots+\frac{1}{\sqrt{99+70 \sqrt{2}}} \\ = & \frac{1}{\sqrt{1+2 \sqrt{2}+2}}+\frac{1}{\sqrt{2+2 \sqrt{2} \cdot \sqrt{3}+3}}+\frac{1}{\sqrt{3+2 \sqrt{3} \cdot \sqrt{4}+4}}+\cdots+\frac{1}{\sqrt{49+2 \cdot \sqrt{49} \cdot \sqrt{50}+50}} \\ = & \frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots+\frac{1}{\sqrt{49}+\sqrt{50}} \\ = & \sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\cdots+\sqrt{50}-\sqrt{49} \\ = & \sqrt{50}-1 \\ = & 5 \sqrt{2}-1 \end{aligned} $$
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最后修改于4月8日
下一篇:问题267: 中学Form 2第2课。可以教我怎么做算式吗?