Question

-3x²+6x＞2怎么解

用2种解法 ~ 1.求根公式
2.配方进行解答

Answer 1

-3x^2+6x>2
-3x^2+6x-2>0
-3(x^2-2x+1)+3-2>0
-3(x-1)^2>-1
(x-1)^2<1/3
-根号3/3<x-1<根号3/3
-根号3/3+1<x<根号3/3+1

-3x^2+6x-2>0
判别式△=36-4（-3）（-2）=36-24=12>0
求根公式得x1=[-6+根号12]/[-3x2]=1+根号3/3
x2=[-6-根号12]/[-3x2]=1-根号3/3
所以范围为 -根号3/3+1<x<根号3/3+1

Answer 2

3x²-6x<-2;
3(x²-2x+1)<1;
(x+1)²<1/3;
-√（1/3)<x+1<√(1/3);
-1-√（1/3)<x<-1+√(1/3)

Answer 3

3x^2-6x+2<0
用求根公式得两个根(3±sqr(3))/3
所以(x-(3+sqr(3))/3)(x-(3-sqr(3))/3))<0
x∈((3-sqr(3))/3,(3+sqr(3))/3)

3x^2-6x+2<0
3(x-1)^2<1
-sqr3/3<x-1<sqr3/3
x∈((3-sqr(3))/3,(3+sqr(3))/3)

Answer 4

很抱歉..解不出