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用分解因式法、适当的方法解方程

2025-04-07 21:45:28

推荐回答（2个）

回答1：

（1）（2x-5）²-（x-2）²=0

[(2x-5)-(x-2)][(2x-5)+(x-2)]=0
(x-3)(3x-7)=0
x=3或x=7/3
（2）x²+（√5+√3）x+√15=0

(x+√5)(x+√3)=0
x=-√5或x=-√3
（3）x²-16x+60=0

(x-6)(x-10)=0
x=6或x=10
（4）(x²+x)(x²+x-2)=24
(x²+x)²-2(x²+x)-24=0
[(x²+x)-6][(x²+x)+4]=0
∵x²+x+4=(x+½)²+15/4＞0
∴x²+x-6=0
(x-2)(x+3)=0
x=2或x=-3
（5）9/4（2x+3）²=25/36（12x-18）²
[3/2(2x+3)]²-[5/6(12x-18)]²=0
[(3x+9/2)-(10x-15)][(3x+9/2)+(10x-15)]=0
(-7x+39/2)(13x-21/2)=0
x=39/14或x=21/26
（6）（x-√2）=√5x（2-√2x）

(x-√2)+√10x(x-√2)=0
(x-√2)(1+√10x)=0
x=√2或x=-(√10)/10
（7）（x+1）（x+2）=2x+4
(x+1)(x+2)-2(x+2)=0
(x+2)(x+1-2)=0
x=-2或x=1
（8）（y+2)²=(3y-1)²

（y+2)²-(3y-1)²=0
[(y+2)-(3y-1)][(y+2)+(3y-1)]=0
(-2y+3)(4y+1)=0
y=1.5或y=-0.25
（1）25x²-144=0
(5x)²=12²
5x=±12
x=±2.4
（2）3x（1-x）=2x-2
3x(x-1)+2(x-1)=0
(3x+2)(x-1)=0
x=-2/3或x=1
（3）x²+3x+2=0
(x+1)(x+2)=0
x=-1或x=-2
（4）√2x²+3x-2√2=0
(√2x-1)(x+2√2)=0
x=√2/2或x=-2√2
（5）（√2-1）x-2x²=0
[(√2-1)-2x]x=0
x=0或x=(√2-1)/2
（6）（√3+√2）x²+2√3x+（√3-√2）=0

[(√3+√2)x+(√3-√2)](x+1)=0
x=-1或x=(√3-√2)/(√3+√2)=(√3-√2)²/(√3+√2)(√3-√2)=7-2√6
（7）（2m+1）²+3（2m+1）+2=0
[(2m+1)+2][(2m+1)+1]=0
2(2m+3)(m+1)=0
m=-1或m=-1.5
(8)x²-4x+1=0
(x-2)²=3
x-2=±√3
x=2±√3
(9)（x-1）(x+2)=70
x²+x-2=70
x²+x-72=0
(x-8)(x+9)=0
x=8或x=-9
(10)9x²=4(3x+1)
9x²-12x-4=0
(3x-2)²=8
3x-2=±2√2
x=(2±2√2)/3
(11)2x²-7x+2=0
2(x-7/4)²=33/8
x-7/4=±√33/4
x=(7±√33)/4
(12)(4x-1)²-10(4x-1)-24=0
[(4x-1)-12][(4x-1)+2]=0
(4x-13)(4x+1)=0
x=13/4或x=-0.25
(13)x^4-x²-6=0
(x²-3)(x²+2)=0
∵ x²+2>0
∴x²=3；x=±√3
(14)x²-2√5x+2=0
(x-√5)²=3
x-√5=±√3
x=√5±√3
(15)3x(1-x)=2x-2
3x(x-1)+2(x-1)=0
(3x+2)(x-1)=0
x=1或x=-1.5

回答2：

先占位
在答
1 (2x-5)^2-(x-2)^2=0
[(2x-5)+(x-2)][(2x-5)-(x-2)]=0
(3x-7)(x-3)=0
x1=7/3, x2=3
2 x^2+(根号5+根号3）x+根号15=0
(x+√5)(x+√3)=0
x1=-√5, x2=-√3
3 x^2-16x+60=0
x²-16x+64=4
(x-8)²=4
x-8=±2
x1=10, x2=6
4 (x^2+x)(x^2+x-2)=24
(x²+x)²-2(x²+x)-24=0
(x²+x-6)(x²+x+4)=0
x²+x-6=0
(x+3)(x-2)=0
x1=-3, x2=2

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