16(x-1)²=9(2x+3)²

问题描述:

16(x-1)²=9(2x+3)²

是解方程吗?移项,平方差公式得 [4(x-1)+3(2x+3)][4(x-1)-3(2x+3)]=0 合并同类项得(10x+5)(2X+13)=0 得 x=-1/2或x=-13/2

16(x-1)²=9(2x+3)²
16(x-1)²-9(2x+3)²=0
[4(x-1)+3(2x+3)][4(x-1)-3(2x+3)]=0
(10x+5)(-2x-13)=0
-5(2x+1)(2x+13)=0
(2x+1)(2x+13)=0
x1=-1/2,x2=-13/2

16(x²-2x+1)=9(4x²+12x+9)
16x²-32x+16=36x²+108x+81
20x²+140x+65=0
4x²+28x+13=0
(2x+1)(2x+13)=0
x1=-2分之1,x2=-2分之13