Solve the inequality about x 1 m x 2 2m x m 3 0, where m R.

1) When m=1, it can be reduced to 2x-4>0, and the inequality is solved as x>2;

2) When m≠1, =4m -4(1-m)[-m+3)]=4(3-2m).

1-m>0 and <0, the inequality is solved as any real number, at which point m<1 and m>, but such m does not exist;

When 1-m>0 and =0, y=(1-m)x 2+2mx-(m+3) has only one intersection point with the x-axis, and the original inequality has a solution, but such m does not exist;

1-m>0 and >0, m<1 and m<, that is, m<1 y=(1-m) x 2+2mx-(m+3) has two intersections with the x-axis, and the original inequality is solved.

x>[-m+ (3-2m)] (1-m) or x<[-m- (3-2m)] (1-m);

1-m<0 and 0, that is, there is no solution to the original inequality at m;

1-m<0 and >0, i.e., 1[-m+ (3-2m)] (1-m) or x<[-m- (3-2m)] (1-m);

When m=1, the inequality is solved as x>2;

When 1 is m, the original inequality has no solution.

The discriminant formula of the quadratic equation of 1-m) x 2+2mx-(m+3)=0 = (2m) 2+4(1-m)(m+3)=12-8m

When 12-8m<0 is m>3 2, (1-m) <0, then (1-m) x 2+2mx-(m+3) is always less than 0, and the equation has no solution.

When 12-8m>=0, that is, m<=3, 2 is divided into three cases.

1-m)<0 1[-m+root(3-2m)] (1-m)x>21-m)>0 m<1 =>x>[-m+root(3-2m)] (1-m) or x<[-m-root(3-2m)] (1-m).

The solution of inequality $(m-2) x 2-2(m-2)x-1>0$ depends on the value of the parameter m. In order to solve for this inequality, we need to first determine the range of m.

First, let's simplify the inequality:

m-2)x^2-2(m-2)x-1>0$

x^2-2x-\frac>0$

This is a quadratic inequality, so we need to find its root first:

x=\frac}}$

In order for the inequality to have a solution, we need to make sure that $ sqrt}$ is a real number where the root number is opened.

This requires $4+ frac>0$, which is $m>2$.

So, when $m>2$, the inequality $(m-2)x 2-2(m-2)x-1>0$ is solved.

Next, we need to determine the solution of the inequality:

x=\frac}}$

This is a pair of inequality roots, and we need to determine the position of these two roots in the unequal finch bridge state:

x= frac}} $If $ frac}}>0$, then the solution of the inequality is $(0,+ infty)$; Otherwise, the solution of the inequality is $(frac}},infty)$.

x= frac}} $If $ frac}}>0$, then the eclipse inequality is enormous. Hope.

x^2-(3m+1)x+2m^2+m>0

x-m)[x-(2m+1)]>0

The two zeros are m and 2m+1

So it's better than laughing jujube grinding than their size rock tease.

m> bumper 2m+1, m0, x+1 is not equal to 0

In summary. m-1,x 2m+1

About x's bumper is unequal to the finch noisy slippery (m+3) x 2+2mx+m-2>0,

m=-3 becomes -6x-5>0, and x-3 4=m 2-(m+3)(m-2)=-m+6,3

Inequality 1 3(x-m) is greater than 2-m and the solution set is that x limb is at 2, then the value of m is ———

1 3 (x-m) is greater than 2-m

1/3)x-(1/3)m>2-m

1 Hunger grinding wide bucket 3) x > 2-(2 3) m

x>6-2m

x>26-2m=2m=2

When the m trace is matched with 2

m (1-m) Zibi refers to (m-2).

When m 2m (1-m) (m-2).

When m=2 there is no solution.

, it is easy to get x>1;

When it is not equal to -3, [(m+3)x-m](x-1)>0, x1=m (m+3)=1-3 (m+3), x2=1,a) m (-3,-2), m+3>0 and x1<0x2, so x (1,m (m+3));

In summary, m<-3, x (1, m (m+3)); m=-3, x>1;-3< m<-2, x (negative infinity, m (m+3)) and (1, positive infinity).

Extract m separately to yield: m 3x (x-1).

Because m r and m <-2

So 0 (3x-2)(x-1).

The answer can be calculated.

((m+3)x-m)(x-1)>0

31, x>m (m+3) or x<1m=-3 is primitively 3x-3>0 then x>1

m<-3 does it give m (m+3).

Using the cross multiplication method, the quadratic trinomial formula is first decomposed, and then the m is classified and discussed.

m+3)x²+(m+2)x-1>0

m+3)x-1](x+1)>0

1) m+3=0, m=-3.

The solution of the inequality is, x<-1

2) m+3<0 and 1 (m+3)<-1,-4-1,m<-4.

The inequality is solved as x<-1 or x>1 (m+3)(5)m+3>0 and 1 (m+3)<-1, which does not exist.

6) m+3>0 and 1 (m+3)>-1, m>-3.

The solution of the inequality is x<-1 or x>1 (m+3).

Divide the situation and collapse:

-2<0, established, x sedan clan r

No=>0, the opening is up, as long as >0, go to the middle part.

4m 2-4m 2*(-2)=12m2 constant "0.".The solution of m 2 x 2 + 2 mx-2 = 0 two roots, x1, x2, is denoted by m.

So. x1