《2026年练就优等生课后提分攻略八年级数学全一册人教版》

01 观察式子: $\frac{1}{1×2}=1-\frac{1}{2},\frac{1}{2×3}=\frac{1}{2}-\frac{1}{3},\frac{1}{3×4}=\frac{1}{3}-\frac{1}{4},...$.
(1)由此得到$\frac{1}{n(n + 1)}=$\frac{1}{n}-\frac{1}{n + 1};
(2)利用上面的结论化简:$\frac{1}{x(x + 1)}+\frac{1}{(x + 1)(x + 2)}+...+\frac{1}{(x + 999)(x + 1000)}$;
(3)化简:$\frac{1}{x(x + 2)}+\frac{1}{(x + 2)(x + 4)}+\frac{1}{(x + 4)(x + 6)}+...+\frac{1}{(x + 2008)(x + 2010)}$.

答案：
$(1)\frac{1}{n}-\frac{1}{n + 1}$　　
$(2) $原式$=\frac{1}{x}-\frac{1}{x + 1}+\frac{1}{x + 1}-\frac{1}{x + 2}+·s+\frac{1}{x + 999}-\frac{1}{x + 1000}=\frac{1}{x}-\frac{1}{x + 1000}=\frac{1000}{x(x + 1000)}$　　
$(3) $原式$=\frac{1}{2}(\frac{1}{x}-\frac{1}{x + 2}+\frac{1}{x + 2}-\frac{1}{x + 4}+·s+\frac{1}{x + 2008}-\frac{1}{x + 2010})=\frac{1}{2}(\frac{1}{x}-\frac{1}{x + 2010})=\frac{1005}{x(x + 2010)}$　　

02 某同学在学习过程中,遇到这样一个问题:“求$A = 48×(\frac{1}{3^{2}-4}+\frac{1}{4^{2}-4}+...+\frac{1}{100^{2}-4})$的整数部分.”他百思不得其解,于是向数学老师请教.数学老师进行了深入浅出地讲解:观察算式中每个分母中的减数都是4,且被减数都在递增.
先看一般情形:$\frac{1}{a^{2}-4}=\frac{1}{4}(\frac{1}{a - 2}-\frac{1}{a + 2})$.
再看特殊情形:
当$a = 3$时,$\frac{1}{3^{2}-4}=\frac{1}{4}×(\frac{1}{3 - 2}-\frac{1}{3 + 2})$;当$a = 4$时,$\frac{1}{4^{2}-4}=\frac{1}{4}×(\frac{1}{4 - 2}-\frac{1}{4 + 2})$.
老师讲解到这里时,该同学说:“老师,我知道怎么做了!”
(1)请你通过化简,说明一般情形$\frac{1}{4}(\frac{1}{a - 2}-\frac{1}{a + 2})=\frac{1}{a^{2}-4}$的正确性;
(2)请你完成该题的解答.

答案：
(1) $\because$ 左边$=\frac{1}{4}×\frac{a + 2}{(a + 2)(a - 2)}-\frac{1}{4}×\frac{a - 2}{(a + 2)(a - 2)}=\frac{1}{4}×\frac{a + 2 - a + 2}{(a + 2)(a - 2)}=\frac{1}{4}×\frac{4}{a^{2}-4}=\frac{1}{a^{2}-4}$，
$\therefore$左边$=$右边，即原等式成立.
(2) $\because\frac{1}{a^{2}-4}=\frac{1}{4}(\frac{1}{a - 2}-\frac{1}{a + 2})$，
$\therefore A = 48×\frac{1}{4}[(1+\frac{1}{2}+·s+\frac{1}{98})-(\frac{1}{5}+\frac{1}{6}+·s+\frac{1}{102})]$
$=12×(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{99}-\frac{1}{100}-\frac{1}{101}-\frac{1}{102})$
$=25 - 12×(\frac{1}{99}+\frac{1}{100}+\frac{1}{101}+\frac{1}{102})$
$\because\frac{1}{100},\frac{1}{101},\frac{1}{102}$都小于$\frac{1}{99}$，
$\therefore 12×(\frac{1}{99}+\frac{1}{100}+\frac{1}{101}+\frac{1}{102})＜12×\frac{4}{99}＜\frac{1}{2}$
$\therefore A$的整数部分是$24$.