《2025年拔尖特训七年级数学上册人教版》

4. 观察下列各式：
$\frac {1}{1×2}= 1-\frac {1}{2},\frac {1}{2×3}= \frac {1}{2}-\frac {1}{3},\frac {1}{3×4}= \frac {1}{3}-\frac {1}{4},$
$\frac {1}{4×5}= \frac {1}{4}-\frac {1}{5}……$
探索规律,根据规律解答下列问题：
(1)第6个等式为$\frac {1}{6×7}$=$\frac {1}{6}-\frac {1}{7}$.
(2)计算：$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+... +\frac {1}{2025×2026}.$
$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+... +\frac {1}{2025×2026}=1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+\frac {1}{3}-\frac {1}{4}+... +\frac {1}{2025}-\frac {1}{2026}=1-\frac {1}{2026}=\frac {2025}{2026}$
(3)若有理数a,b满足$|a-3|+|b-5|= 0,$
求$\frac {1}{ab}+\frac {1}{(a+2)(b+2)}+\frac {1}{(a+4)(b+4)}+... +\frac {1}{(a+100)(b+100)}$的值.
因为$|a-3|+|b-5|=0$，所以$a-3=0,b-5=0$.所以$a=3,b=5$.所以$\frac {1}{ab}+\frac {1}{(a+2)(b+2)}+\frac {1}{(a+4)(b+4)}+... +\frac {1}{(a+100)(b+100)}=\frac {1}{3×5}+\frac {1}{5×7}+\frac {1}{7×9}+... +\frac {1}{103×105}=\frac {1}{2}×(\frac {1}{3}-\frac {1}{5}+\frac {1}{5}-\frac {1}{7}+\frac {1}{7}-\frac {1}{9}+... +\frac {1}{103}-\frac {1}{105})=\frac {1}{2}×(\frac {1}{3}-\frac {1}{105})=\frac {1}{2}×\frac {105-3}{3×105}=\frac {1}{2}×\frac {102}{315}=\frac {17}{105}$

答案： 4.（1）$\frac {1}{6×7};\frac {1}{6}-\frac {1}{7}$.
（2）$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+... +\frac {1}{2025×2026}=1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+\frac {1}{3}-\frac {1}{4}+... +\frac {1}{2025}-\frac {1}{2026}=1-\frac {1}{2026}=\frac {2025}{2026}$.
（3）因为$|a-3|+|b-5|=0$，所以$a-3=0,b-5=0$.所以$a=3,b=5$.所以$\frac {1}{ab}+\frac {1}{(a+2)(b+2)}+\frac {1}{(a+4)(b+4)}+... +\frac {1}{(a+100)(b+100)}=\frac {1}{3×5}+\frac {1}{5×7}+\frac {1}{7×9}+... +\frac {1}{103×105}=\frac {1}{2}×(\frac {1}{3}-\frac {1}{5}+\frac {1}{5}-\frac {1}{7}+\frac {1}{7}-\frac {1}{9}+... +\frac {1}{103}-\frac {1}{105})=\frac {1}{2}×(\frac {1}{3}-\frac {1}{105})=\frac {1}{2}×\frac {105-3}{3×105}=\frac {1}{2}×\frac {102}{315}=\frac {17}{105}$.

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+…[phone].
(3)$|\frac {1}{2}-1|+|\frac {1}{3}-\frac {1}{2}|+|\frac {1}{4}-\frac {1}{3}|+... +|\frac {1}{2026}-\frac {1}{2025}|.$

答案： 5.（1）原式=(1-2)+(3-4)+(5-6)+…+(99-100)=(-1)+(-1)+(-1)+…+(-1)=(-1)×50=-50.
（2）原式=(2-4-6+8)+(10-12-14+16)+(18-20-22+24)+…+(2018-2020-2022+2024)+(2026-2028)=0+0+0+…+0+(-2)=-2.
（3）原式=$1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+\frac {1}{3}-\frac {1}{4}+... +\frac {1}{2025}-\frac {1}{2026}=1-\frac {1}{2026}=\frac {2025}{2026}$.