Question

(3x-1)的平方=1 2(x+1)的平方=x的平方-1 (2x-1)的平方+2(2x-1)=3 (y+3)(1-3y)=1+2y的平方

用适当方法解下列方程

Answer 1

(3x-1)²=1 3x-1=±1 3x=1±1
3x=2 or 3x=0
x=2/3 or x=0

2(x+1)²=x²-1
2(x²+2x+1)=x²-1
2x²+4x+2=x²-1
x²+4x+3=0
(x+1)(x+3)=0
x=-1 or x=-3

(2x-1)²+2(2x-1)=3
4x²-4x+1+4x-2=3
4x²=4
x²=1
x=1 or x=-1

(y+3)(1-3y)=(1+2y)²
-3y²-9y+y+3=1+4y+4y²
7y²+12y-2=0
y=-12±√(12²+4*2*7) /7*2
=-6±√50 /7

Answer 2

(3x-1)^2=1
3x-1=±1
x1=(1+1)/3=2/3
x2=(1-1)/3=0

2(x+1)^2=x^2-1
2(x+1)^2=(x+1)(x-1)
(x+1)［2(x+1)-(x-1)］=0
(x+1)(2x+2-x+1)=0
(x+1)(x+3)=0
x1=-1
x2=-3

(2x-1)^2+2(2x-1)=3
(2x-1)^2+2(2x-1)-3=0
(2x-1+3)(2x-1-1)=0
(2x+2)(2x-2)=0
4(x+1)(x-1)=0
x1=-1
x2=1

(y+3)(1-3y)=1+2y^2
y-3y^2+3-9y=1+2y^2
5y^2+8y-2=0
y^2+8/5*y-2/5=0
y^2+8/5*y+16/25-16/25-2/5=0
(y+4/5)^2=26/25
y+4/5=±√26/5
y1=√26/5-4/5=(√26-4)/5
y2=-√26/5-4/5=(-√26-4)/5