《2025年暑假总动员宁夏人民教育出版社七年级数学沪科版》

17. 先观察下列等式,然后用你发现的规律解答下列问题.
$\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)计算$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+\frac {1}{4×5}+\frac {1}{5×6}= $$\frac{5}{6}$.
(2)探究$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+... +\frac {1}{n×(n+1)}= $$\frac{n}{n + 1}$.(用含有n的式子表示)
(3)若$\frac {1}{1×3}+\frac {1}{3×5}+\frac {1}{5×7}+... +\frac {1}{(2n-1)×(2n+1)}$的值为$\frac {17}{35}$,求n的值.
解：$\frac {1}{1×3}$ + $\frac {1}{3×5}$ + $\frac {1}{5×7}$ +... + $\frac {1}{(2n - 1)×(2n + 1)}$ = $\frac{1}{2}$×(1 - $\frac{1}{3}$) + $\frac{1}{2}$×($\frac{1}{3}$ - $\frac{1}{5}$) + $\frac{1}{2}$×($\frac{1}{5}$ - $\frac{1}{7}$) +... + $\frac{1}{2}$×($\frac{1}{2n - 1}$ - $\frac{1}{2n + 1}$) = $\frac{1}{2}$×(1 - $\frac{1}{2n + 1}$) = $\frac{n}{2n + 1}$,由$\frac{n}{2n + 1}$ = $\frac{17}{35}$,解得n = 17.经检验，n = 17是方程的根,所以n = 17.

答案：
(1) $\frac{5}{6}$　
(2) $\frac{n}{n + 1}$
(3) $\frac{1}{1×3}$ + $\frac{1}{3×5}$ + $\frac{1}{5×7}$ +... + $\frac{1}{(2n - 1)×(2n + 1)}$ = $\frac{1}{2}$×(1 - $\frac{1}{3}$) + $\frac{1}{2}$×($\frac{1}{3}$ - $\frac{1}{5}$) + $\frac{1}{2}$×($\frac{1}{5}$ - $\frac{1}{7}$) +... + $\frac{1}{2}$×($\frac{1}{2n - 1}$ - $\frac{1}{2n + 1}$) = $\frac{1}{2}$×(1 - $\frac{1}{2n + 1}$) = $\frac{n}{2n + 1}$,由$\frac{n}{2n + 1}$ = $\frac{17}{35}$,解得n = 17.经检验，n = 17是方程的根,所以n = 17.

18. 定义“⊕”为一种新的运算符号,先观察下列各式：
$1\oplus 3= 1×3-3= 0$；
$-4\oplus 5= (-4)×3-5= -17$；
$2\oplus (-\frac {1}{3})= 2×3-(-\frac {1}{3})= 6\frac {1}{3}$；
$0\oplus 6= 0×3-6= -6$；
$-\frac {1}{4}\oplus (-4)= -\frac {1}{4}×3-(-4)= 3\frac {1}{4}$；……
(1)根据以上算式,写出$a\oplus b=$3a - b.
(2)根据(1)中定义的$a\oplus b$的运算规则,解决下面问题：
①若$x= 4$,求$(x-2)\oplus 4x$的值；
②若$2m-n= -2$,求$(m+n)\oplus (-5m+7n)$的值.
(2) ① $(x - 2) ⊕ 4x = 3(x - 2) - 4x = -x - 6$,
当x = 4时, $(x - 2) ⊕ 4x$=-10.
② $(m + n) ⊕ (-5m + 7n) = 3(m + n) - (-5m + 7n) = 3m + 3n + 5m - 7n = 8m - 4n = 4(2m - n)$,
当2m - n = -2时,原式=-8.

答案：
(1) 3a - b
(2) ① (x - 2) ⊕ 4x = 3(x - 2) - 4x = -x - 6,
当x = 4时, (x - 2) ⊕ 4x = -4 - 6 = -10.
② (m + n) ⊕ (-5m + 7n) = 3(m + n) - (-5m + 7n) = 3m + 3n + 5m - 7n = 8m - 4n = 4(2m - n),
当2m - n = -2时,原式 = 4×(-2) = -8.