用因式分解法解x的平方-2x=0
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(2x-1)²=(3-x)²(2x-1)²-(3-x)²=0[(2x-1)+(3-x)][(2x-1)-(3-x)]=0(x+2)(3x-4)=0x+2=0或3
解(x-3)²+2x(x-3)=0(x-3)(x-3+2x)=0(x-3)(3x-3)=0∴x=3或x=1再问:应该用提公因式法?再答:提了提了(x-3)∴(x-3)²+2x(x-
解(1)因式分解发法由(2x-1)²=(3-x)²即(2x-1)²-(3-x)²=0即(2x-1+3-x)(2x-1-3+x)=0即(x+2)(3x-4)=0即
(2x+1)²=(1-3x)²(2x+1)²-(1-3x)²=0(2x+1+1-3x)(2x+1-1+3x)=05x(2-x)=0x=0或x=2.
(3x-1)^-【2(2x+3)】^=0(3x-1+4x+6)(3x-1-4x-6)=0(7x+5)(-x-7)=0x1=-5/7,x2=-7
(x+2)^2=(5-2x)^2(x+2)^2-(5-2x)^2=0(x+2+5-2x)(x+2-5+2x)=0(7-x)(3x-3)=0x=7或1
(2x-1)^2=(3-x)^2(2x-1)^2-(3-x)^2=0[(2x-1)+(3-x)]*[(2x-1)-(3-x)]=0(2x-1)+(3-x)=0或(2x-1)-(3-x)=0x=-2或x
(x-3)²=2x(3-x)(3-x)²=2x(3-x)(3-x-2x)(3-x)=0(3-3x)(3-x)=0x1=1,x2=3
x²-2x-8=0(x-1)²-3²=0(x-4)(x+2)=0x1=4,x2=-2
(x-4)^2=(5-2x)^2(x-4)^2-(5-2x)^2=0移项(x-4+5-2x)(x-4-5+2x)=0平方差公式(1-x)(3x-9)=0合并同类项3(x-3)(1-x)=0提取最大公因
简单化简为x²-5x-24=0再化为(x+3)(x-8)=0解得x=-3或x=8
解(2x+1)²=(x-3)²(2x+1)²-(x-3)²=0[(2x+1)+(x-3)][(2x+1)-(x-3)]=0(3x-2)(x+4)=0∴x=-4或
(2x+1)的平方=4(3x-2)的平方(2x+1)的平方-4(3x-2)的平方=0[2x+1+2(3x-2)][2x+1-2(3x-2)]=0[8x-3][-4x+5]=0x=3/8x=5/4
(x-2)=4或x-2=-4=>x1=6,x2=-2
:(2x-3)的平方-(3x-2)的平方=0[(2x-3)+(3x-2)][(2x-3)-(3x-2)]=0[5x-5][-x-1]=05x-5=0x=1-x-1=0x=-1
x^2-x-2=0(x-2)(+1)=0x=2或x=-1这种因式分解最好是用十字相乘法下面是这方面的资料
3(x²-1)=(x-1)²3x²-3=x²-2x+12x²+2x-4=0x²+x-2=0(x-1)(x+2)=0x1=1,x2=-2手机提
x的平方-6x+9=(1-2x)的平方x²-6x+9=1-4x+4x²3x²-2x-8=0十字相乘法:1-2×34(x-2)(3x+4)=0x-2=0得:x=23x+4=
x²-6x+9=(5-2x)²x²-6x+9=25+4x²-20x3x²-14x+16=0(x-2)(3x-8)=0x1=2x2=8/3~