(x-1)(x+3)=12,(3x-5)(2x-1)= -12x+7,(6x+5)(6y-5)-24=0,4(x-3)&

(x-1)(x+3)=12,(3x-5)(2x-1)= -12x+7,(6x+5)(6y-5)-24=0,4(x-3)²-x(x-3)=0用因式分解怎么解?
品行很好 1年前 已收到2个回答 举报

歪叽呀呀 幼苗

共回答了16个问题采纳率:100% 举报

(x-1)(x+3)=12
x^2 + 2x - 3 = 12 ,x^2 + 2x - 15 = 0 = (x+5)(x-3)
x = 5 ,x = 3
(3x-5)(2x-1)= -12x+7 = 6x^2 - 13x + 5
6x^2 - x - 2 = 0 = (2x + 1)(3x - 2)
x = -1/2 ,x = 2/3
(6x+5)(6x-5)-24=0 = 36x^2 - 49 = (6x + 7)(6x - 7)
x = -7/6 ,x = 7/6
4(x-3)2-x(x-3)=0 = (x - 3)[4(x - 3) - x] = 3(x - 3)(x - 4)
x = 3 ,x = 4

1年前

8

skcarson 幼苗

共回答了293个问题 举报

都化成x的齐次方+常数=0的形式,再因式分解即可。

1年前

2
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