初三数学题，用因式分解法解一元二次方程。 初三数学如何用因式分解法解一元二次方程
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x^2-3x+2=0
(x-1)(x-2)=0
x=1 x=2
2. (x-3)^2-2(x-3)+1=0
[(x-3)-1]^2=0
x-4=0
x=4
3. (3x-4)^2=9x-12
(3x-4)^2=3(3x-4)
(3x-4\uff09\uff083x-4-3)=0
\uff083x-4)\uff083x-7)=0
x=4/3 x=7/3
4. 9(2x+3)^2-4(2x-5)^2=0
[3(2x+3)+2(2x-5)][3(2x+3)-2(2x-5)]=0
(10x-1)(2x+19)=0
x=1/10 x=-19/2
5. \u82e5\u6574\u5f0f 4x^2-2x-5\u4e0e2x^+1\u4e92\u4e3a\u76f8\u53cd\u6570 \u6c42x\u7684\u503c
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\u4e8e\u662f\u67094x^2-2x-5+2x^2+1=0
6x^2-2x-4=0
3x^2-x-2=0
(3x+2)(x-1)=0
x=-2/3 x=1
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(14) (x+1)(x+2)(x+3)(x+4)=24[(x+1)(x+4)][(x+2)(x+3)]=24
(x²+5x+4)(x²+5x+6)=24
(x²+5x)²+10(x²+5x)+24=24
(x²+5x)²+10(x²+5x)=0
(x²+5x+10)(x²+5x)=0
而x²+5x+10=(x+5/2)²+15/4>0恒成立
∴x²+5x=0 ∴解得x1=0，x2=-5
(12)x²-2xy-3y²=0
(x-3y)(x+y)=0
∴x=3y或x=-y
当x=3y时，(2x+3y)/(3y)=3
当x=-y时，(2x+3y)/(3y)=1/3
(13)∵BC、AC是关于x的一元二次方程x²-2(m-1)x+m(m-2)=0的两个根
∴由根与系数关系可知BC+AC=2(m-1) BC·AC=m(m-2)
又AB为Rt△ABC的斜边，BC、AC为直角边
∴AB²=BC²+AC²=(BC+AC)²-2BC·AC=4(m-1)²-2m(m-2)=2m²+4m+4=10
∴m²+2m-3=0 解得m=1或m=-3
而△=[2(m-1)]²-4m(m-2)=4
故满足题意的m值为-3或1
(x+1)(x+2)(x+3)(x+4) = 24
f(x) =(x+1)(x+2)(x+3)(x+4) - 24
f(0) =0
f(-5) = 0
f(x) =(x+1)(x+2)(x+3)(x+4) - 24 = x(x+5)(x^2+ax +b)
x=-1
-24 = -4(1-a +b)
-a+b = 5 (1)
x=-2
-24=-6(4-2a +b)
-2a+b =0 (2)
(1)-(2)
a=5
from (1)
-5+b=5
b=10
f(x) =(x+1)(x+2)(x+3)(x+4) - 24 = x(x+5)(x^2+5x +10)
=>
(x+1)(x+2)(x+3)(x+4) = 24
x(x+5)(x^2+5x +10) =0
x=0 or -5
(13)
x^2-2xy-3y^2=0
(x-3y)(x+y) =0
x=3y or -y
case 1: x=3y
(2x+3y)/(3y)
=9y/(3y)
=3
case 2: x=-y
(2x+3y)/(3y)
=y/(3y)
=1/3
ie
(2x+3y)/(3y) = 3 or 1/3
(13)
AB=10
x1, x2 roots of equation
x^2-2(m-1)x+m(m-2)=0
x1+x2= 2(m-1)
x1.x2= m(m-2)
|AB|^2 = |BC|^2 + |AC|^2
(x1)^2 +(x2)^2=100
(x1+x2)^2 -2x1x2 =100
4(m-1)^2-2m(m-2) =100
2m^2-4m+4 =100
m^2 -2m- 48 =0
(m+6)(m-8)=0
m=8 or -6
x²-2xy-3y²=（x-3y）（x+y）=0
则x=3y或x=-y
所以（2x+3y）/3y=3或1/3
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