初二下学期数学分式

来源：百度知道 编辑：UC知道 时间：2024/06/29 02:14:10
① (a-2) a-1 a-4
{ ----- - -------- } 除以 -----
a的2次方+2a a的2次方+4a+4 a+2

② x的2次方-1 B C
------------ = A+ --- + ---
(x-2)(x-3) x-2 x-3 求 A.B.C

1
{ - --- } 的-2次方
4

[(a-2)/(a^2+2a)-(a-1)/(a^2+4a+4)]/[(a-4)/(a+2)]
=[(a-2)/a(a+2)-(a-1)/(a+2)^2]/[(a-4)/(a+2)]
={[(a-2)(a+2)-a(a-1)]/a(a+2)^2]*[(a+2)/(a-4)]
=(a^2-4-a^2+a)(a+2)/a(a+2)^2(a-4)
=1/a(a+2)
=1/(a^2+2a)

A+B/(x-2)+C/(x-3)
=[A(x-2)(x-3)+B(x-3)+C(x-2)]/(x-2)(x-3)
=[Ax^2+(-5A+B+C)x+(6A-3B-2C)]/(x-2)(x-3)
=(x^2-1)/(x-2)(x-3)
所以A=1
-5A+B+C=0
6A-3B-2C=-1
所以A=1,B=-3,C=8

(1/4)^(-2)
=[(1/4)^(-1)]^2
=4^2
=16