《2025年拔尖特训七年级数学上册人教版》

典例1 计算:$(\frac {1}{2}+\frac {1}{3}+... +\frac {1}{2026})×(1+\frac {1}{2}+$
$\frac {1}{3}+... +\frac {1}{2025})-(1+\frac {1}{2}+\frac {1}{3}+... +\frac {1}{2026})×$
$(\frac {1}{2}+\frac {1}{3}+... +\frac {1}{2025}).$

答案： 设$1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{2025}=a$，$\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{2025}=b$，则$a - b = 1$.所以原式$=(b+\frac{1}{2026})a-(a+\frac{1}{2026})b=\frac{a}{2026}-\frac{b}{2026}=\frac{a - b}{2026}=\frac{1}{2026}$.

[变式] 计算:
(1)$1+\frac {1}{1+2}+\frac {1}{1+2+3}+... +\frac {1}{1+2+... +100}.$
(2)$(1-\frac {1}{2}-\frac {1}{3}-\frac {1}{4}-\frac {1}{5})×(\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+$
$\frac {1}{6})-(1-\frac {1}{2}-\frac {1}{3}-\frac {1}{4}-\frac {1}{5}-\frac {1}{6})×(\frac {1}{2}+\frac {1}{3}+$
$\frac {1}{4}+\frac {1}{5}).$

答案：
(1)原式$=2×(\frac{1}{1× 2}+\frac{1}{2× 3}+\frac{1}{3× 4}+\cdots +\frac{1}{100× 101})=2×(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots +\frac{1}{100}-\frac{1}{101})=2×(1-\frac{1}{101})=\frac{200}{101}$.
(2)令$\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=a$，$1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}=b$，则$a + b = 1$.所以原式$=b(a+\frac{1}{6})-(b-\frac{1}{6})a=ab+\frac{1}{6}b - ab+\frac{1}{6}a=\frac{1}{6}(a + b)=\frac{1}{6}× 1=\frac{1}{6}$.

典例2 计算:
(1)$-2^{2}+(-2)^{3}×5-(-\frac {3}{4})÷\frac {1}{4}.$
(2)$-3^{2}-(-1\frac {1}{2})×\frac {2}{3}-6÷|-\frac {1}{3}|^{3}.$

答案：
(1) - 41.
(2) - 170.

[变式] 计算:
(1)$(-2)^{2}×5-(-2)^{3}÷4.$
(2)$3×(-2)^{3}-4×(-3)^{2}+8.$

答案：
(1) 22.
(2) - 52.

典例3 对于有理数a,b,定义新运算"△",规则如下:$a△b= ab-a-b+4$,如$3△5= 3×5-$
$3-5+4= 11.$
(1)求$3△(-4)$的值.
(2)请你判断交换律在"△"运算中是否成立,并说明理由.

答案：
(1)因为$a\triangle b = ab - a - b + 4$，所以$3\triangle(-4)=3×(-4)-3-(-4)+4=-12+(-3)+4+4=-7$；
(2)交换律在“$\triangle$”运算中成立.理由：由题意，可得$a\triangle b = ab - a - b + 4$，$b\triangle a = ab - b - a + 4$，所以$a\triangle b = b\triangle a$.所以交换律在“$\triangle$”运算中成立.