Question

用因式分解法解下列方程

5x(x-1)+2-2x
(2x+1)²=4(2x+1)
2(x+3)²=x²-9
x²+4√3 x=-12
4(x+3)²-9x²=0
要过程啊
(1-a)x²+2x+2=0有实数根，求a的取值范围
不要过程

Answer 1

5x(x-1)+2=2x
5x(x-1)-2（x-1)=0
(x-1)(5x-2)=0
x1=1,x2=2/5

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

2(x+3)^2=x^2-9
2(x+3)^2-(x^2-9)=0
2(x+3)^2-(x+3)(x-3)=0
(x+3)(x+9)=0
x1=-3,x2=-9

x²+4√3 x=-12

x²+4√3 x+12=0
(x+2√3)²=0
x1=x2=-2√3

4(x+3)²-9x²=0
[2(x+3)-3x][2(x+3)+3x]=0
(6-x)(5x+6)=0
x1=6,x2=-6/5

因为 (1-a)x²+2x+2=0有实数根

所以 2^2-4(1-a)*2>=0
解得 a>= - 1/2
又因为 1-a不等于0
a不等于1

所以 a>= - 1/2，且 a不等于1

Answer 2

5x(x-1)+2-2x=0
5x(x-1)-2（x-1)=0
(x-1)(5x-2)=0
x1=1,x2=2/5

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

2(x+3)²=x²-9
2(x+3)²-(x²-9)=0
2(x+3)²-(x+3)(x-3)=0
(x+3)(x+9)=0
x1=-3,x2=-9

x²+4√3 x=-12
x²+4√3 x+12=0
(x+2√3)²=0
x1=x2=-2√3

4(x+3)²-9x²=0
[2(x+3)-3x][2(x+3)+3x]=0
(6-x)(5x+6)=0
x1=6,x2=-6/5

Answer 3

(1-a)x²+2x+2=0有实数根，求a的取值范围
a≥1/2