(2x+1.5)^2=(0.5-x)^2
2x+1.5=+-√(0.5-x)^2
=0.5-x or 2x+1.5=x-0.5
之後應該點做?
1元2次方程(1條)
2007-09-12 6:46 am
回答 (3)
2007-09-12 6:50 am
✔ 最佳答案
( 2x + 1.5 )² = ( 0.5 - x )²( 2x + 1.5 )² - ( 0.5 - x )² = 0
[( 2x + 1.5 ) + ( 0.5 - x )] [( 2x + 1.5 ) - ( 0.5 - x )] = 0
( x + 2 )( 3x + 1 ) = 0
x + 2 = 0 or 3x + 1 = 0
x = - 2 or x = - 1 / 3
2007-09-12 6:58 am
(2x+1.5)^2=(0.5-x)^2
(4x+3)^2=(1-2x)^2
4x+3=1-2x or 4x+3=2x-1
6x= -2 or 2x= -4
x= -1/3~~ or x= -2~~
(4x+3)^2=(1-2x)^2
4x+3=1-2x or 4x+3=2x-1
6x= -2 or 2x= -4
x= -1/3~~ or x= -2~~
2007-09-12 6:54 am
我倒是覺得......應該這樣做
(2x+1.5)^2 = (0.5-x)^2
(2x+1.5)(2x+1.5) = (0.5-x)(0.5-x)
4x^2+6x+2.25 = x^2-x+0.25
3x^2+7x+2 = 0
(3x+1)(x+2) = 0
x = -2 or x = -1/3
(2x+1.5)^2 = (0.5-x)^2
(2x+1.5)(2x+1.5) = (0.5-x)(0.5-x)
4x^2+6x+2.25 = x^2-x+0.25
3x^2+7x+2 = 0
(3x+1)(x+2) = 0
x = -2 or x = -1/3
收錄日期: 2021-04-13 18:53:33
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https://hk.answers.yahoo.com/question/index?qid=20070911000051KK04651