Question

要用因式分解法来解这几个方程.3y^2-6y=0x^2-8x+16=0(1+x)^2-9=0(x+2)(x+3)=(x+

Answer 1

3y²-6y=0

3y(y-2)=0

y=0 或 y-2=0

y1=0

y2=2

x²-8x+16=0

(x-4)²=0

x-4=0

x1=x2=4

(1+x)²-9=0

(1+x)²-3²=0

[(1+x)+3] [(1+x)-3]=0

(x+4)(x-2)=0

x+4=0 或 x-2=0

x1=-4

x2=2

(x+2)(x+3)=(x+3)

(x+2)(x+3)-(x+3)=0

(x+3)[(x+2)-1]=0

(x+3)(x+1)=0

x+3=0 或 x+1=0

x1=-3

x2=-1