678* Вправа 678Алгебра 8 клас Тарасенкова 2025 · § 15 · стор. 197

678. Обчисліть:

${\displaystyle \frac{1}{\sqrt{k}+\sqrt{k+1}}=}$${\displaystyle \frac{\sqrt{k+1}-\sqrt{k}}{(k+1)-k}=}$${\displaystyle \sqrt{k+1}-\sqrt{k}}$.

1) ${\displaystyle \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\dots +\frac{1}{\sqrt{24}+\sqrt{25}}=}$${\displaystyle (\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+\dots +(\sqrt{25}-\sqrt{24})=}$${\displaystyle \sqrt{25}-1=}$${\displaystyle 4}$;

2) ${\displaystyle \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\dots +\frac{1}{\sqrt{99}+\sqrt{100}}=}$${\displaystyle \sqrt{100}-1=}$${\displaystyle 9}$;

${\displaystyle \frac{1}{\sqrt{k}+\sqrt{k+2}}=}$${\displaystyle \frac{\sqrt{k+2}-\sqrt{k}}{2}}$;

3) ${\displaystyle \frac{1}{1+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\dots +\frac{1}{\sqrt{23}+\sqrt{25}}=}$${\displaystyle \frac{(\sqrt{3}-1)+(\sqrt{5}-\sqrt{3})+\dots +(\sqrt{25}-\sqrt{23})}{2}=}$${\displaystyle \frac{5-1}{2}=}$${\displaystyle 2}$;

4) ${\displaystyle \frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+\dots +\frac{1}{\sqrt{70}+\sqrt{72}}=}$${\displaystyle \frac{\sqrt{72}-\sqrt{2}}{2}=}$${\displaystyle \frac{6\sqrt{2}-\sqrt{2}}{2}=}$${\displaystyle \frac{5\sqrt{2}}{2}}$.