4. 观察下列各式:
$ \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}, \frac{1}{4 × 5}= \frac{1}{4}-\frac{1}{5}, … $.
根据规律解答问题:
(1) 第6个等式为
$ \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}, \frac{1}{4 × 5}= \frac{1}{4}-\frac{1}{5}, … $.
根据规律解答问题:
(1) 第6个等式为
$\frac{1}{6×7}$
=$\frac{1}{6} - \frac{1}{7}$
;答案：(1) $\frac{1}{6×7}$ $\frac{1}{6} - \frac{1}{7}$ (2) $\frac{1}{1×2} + \frac{1}{2×3} + \frac{1}{3×4} + \cdots + \frac{1}{2025×2026} = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2025} - \frac{1}{2026} = 1 - \frac{1}{2026} = \frac{2025}{2026}$ (3) 因为 $|a - 3| + |b - 5| = 0$，所以 $a - 3 = 0$，$b - 5 = 0$，解得 $a = 3$，$b = 5$。所以 $\frac{1}{ab} + \frac{1}{(a + 2)(b + 2)} + \frac{1}{(a + 4)(b + 4)} + \cdots + \frac{1}{(a + 100)(b + 100)} = \frac{1}{3×5} + \frac{1}{5×7} + \frac{1}{7×9} + \cdots + \frac{1}{103×105} = \frac{1}{2} × \left(\frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + \cdots + \frac{1}{103} - \frac{1}{105}\right) = \frac{1}{2} × \left(\frac{1}{3} - \frac{1}{105}\right) = \frac{1}{2} × \frac{34}{105} = \frac{17}{105}$
解析：
(1) $\frac{1}{6×7}$ $\frac{1}{6} - \frac{1}{7}$
(2) 计算: $ \frac{1}{1 × 2}+\frac{1}{2 × 3}+\frac{1}{3 × 4}+…+\frac{1}{2025 × 2026} $;
(3) 若有理数 $ a, b $ 满足 $ |a-3|+|b-5|= 0 $, 求 $ \frac{1}{a b}+\frac{1}{(a+2)(b+2)}+\frac{1}{(a+4)(b+4)}+…+ \frac{1}{(a+100)(b+100)} $ 的值.
(3) 若有理数 $ a, b $ 满足 $ |a-3|+|b-5|= 0 $, 求 $ \frac{1}{a b}+\frac{1}{(a+2)(b+2)}+\frac{1}{(a+4)(b+4)}+…+ \frac{1}{(a+100)(b+100)} $ 的值.
答案：(1) $\frac{1}{6×7}$ $\frac{1}{6} - \frac{1}{7}$ (2) $\frac{1}{1×2} + \frac{1}{2×3} + \frac{1}{3×4} + \cdots + \frac{1}{2025×2026} = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2025} - \frac{1}{2026} = 1 - \frac{1}{2026} = \frac{2025}{2026}$ (3) 因为 $|a - 3| + |b - 5| = 0$，所以 $a - 3 = 0$，$b - 5 = 0$，解得 $a = 3$，$b = 5$。所以 $\frac{1}{ab} + \frac{1}{(a + 2)(b + 2)} + \frac{1}{(a + 4)(b + 4)} + \cdots + \frac{1}{(a + 100)(b + 100)} = \frac{1}{3×5} + \frac{1}{5×7} + \frac{1}{7×9} + \cdots + \frac{1}{103×105} = \frac{1}{2} × \left(\frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + \cdots + \frac{1}{103} - \frac{1}{105}\right) = \frac{1}{2} × \left(\frac{1}{3} - \frac{1}{105}\right) = \frac{1}{2} × \frac{34}{105} = \frac{17}{105}$
5. 计算:
(1) $ 1-2+3-4+5-6+…+99-100 $;
(2) $ 2-4-6+8+10-12-14+16+18-20-22+24+…+2010-2012 $;
(3) $ \left|\frac{1}{2}-1\right|+\left|\frac{1}{3}-\frac{1}{2}\right|+\left|\frac{1}{4}-\frac{1}{3}\right|+…+\left|\frac{1}{2025}-\frac{1}{2024}\right| $.
(1) $ 1-2+3-4+5-6+…+99-100 $;
(2) $ 2-4-6+8+10-12-14+16+18-20-22+24+…+2010-2012 $;
(3) $ \left|\frac{1}{2}-1\right|+\left|\frac{1}{3}-\frac{1}{2}\right|+\left|\frac{1}{4}-\frac{1}{3}\right|+…+\left|\frac{1}{2025}-\frac{1}{2024}\right| $.
答案：(1) 原式 $=(1 - 2) + (3 - 4) + (5 - 6) + \cdots + (99 - 100) = (-1) + (-1) + (-1) + \cdots + (-1) = -50$ (2) 原式 $=(2 - 4 - 6 + 8) + (10 - 12 - 14 + 16) + (18 - 20 - 22 + 24) + \cdots + (2002 - 2004 - 2006 + 2008) + (2010 - 2012) = 0 + 0 + 0 + \cdots + 0 + (-2) = -2$ (3) 原式 $=1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2024} - \frac{1}{2025} = 1 - \frac{1}{2025} = \frac{2024}{2025}$