Question

(y+3)(1-3y)=-1+2y^2;(2+根号3)y^2=(2-根号3)y;(3x+2)^2-2x^2=0因式分解解

Answer 1

(y+3)(1-3y)=-1+2y²

3-7y-9y²+1-2y²=0

-11y²-7y+4=0

11y²+7y-4=0

(11y-4)(y+1)=0

11y-4=0 y+1=0

∴y1=4/11

y2=-1

(2+√3)y²=(2-√3)y

2y²+√3y²-2y+√3y=0

y(2y+√3y-2+√3)=0

y=0 2y+√3y=2-√3

∴y1=0

y2=7-4√3

(3x+2)²-2x²=0

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

x+2=0 5x+2=0

∴x1=-2

x2=-2/5