《2025年1加1轻巧夺冠完美期末八年级数学上册人教版辽宁专版》

20. (8分)(葫芦岛绥中期末,有改编)
探索发现:
$\frac{1}{1 × 2} = 1 - \frac{1}{2}$;
$\frac{1}{2 × 3} = \frac{1}{2} - \frac{1}{3}$;
$\frac{1}{3 × 4} = \frac{1}{3} - \frac{1}{4}$;
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根据你发现的规律,回答下列问题.
(1)计算:$\frac{1}{1 × 2} + \frac{1}{2 × 3} + \frac{1}{3 × 4} + ·s + \frac{1}{n(n + 1)}$.
(2)解方程:$\frac{1}{x(x + 1)} + \frac{1}{(x + 1)(x + 2)} + ·s + \frac{1}{(x + 2026)(x + 2027)} = \frac{1}{x + 2027}$.

答案： 20.解：$(1)\frac{1}{1×2} + \frac{1}{2×3} + \frac{1}{3×4} + ⋯ + \frac{1}{n(n + 1)} = 1 - \frac{1}{n + 1} = \frac{n}{n + 1}$
$(2)\frac{1}{x(x + 1)} + \frac{1}{(x + 1)(x + 2)} + ⋯ + \frac{1}{(x + 2026)(x + 2027)} = \frac{1}{x} - \frac{1}{x + 1} + \frac{1}{x + 1} - \frac{1}{x + 2} + ⋯ + \frac{1}{x + 2026} - \frac{1}{x + 2027} = \frac{1}{x} - \frac{1}{x + 2027} = \frac{1}{x + 2027}，$x + 2027 - x = x，x = 2027，经检验，x = 2027是原分式方程的解.