解
(1)x²+2x=x(x+2)
(2)x²+6x+9=(x+3)²
(3)(x-2)²-2(x-2)
=(x-2)[(x-2)-2]
=(x-2)(x-4)
2
3x²-1=0
x²=1/3
∴x=√3/3或x=-√3/3
3
25x²=10x-1
25x²-10x+1=0
(5x-1)²=0
∴x=1/5
解
(1)x²+2x=x(x+2)
(2)x²+6x+9=(x+3)²
(3)(x-2)²-2(x-2)
=(x-2)[(x-2)-2]
=(x-2)(x-4)
2
3x²-1=0
x²=1/3
∴x=√3/3或x=-√3/3
3
25x²=10x-1
25x²-10x+1=0
(5x-1)²=0
∴x=1/5