Mathematics problems about the solution of inequalities...

1) c brings in the value in a.

2)d 3)a<5, divided into x<-3,-3<=x<=2,x>2 The minimum value of x-2 + x+3 is 5, and the solution set of inequality x-2 + x+3 a is empty, so a<5

4) When x -6 x

x│)^2-│x│-6<0

x│-3)(│x│+2)<0

So x -3<0

3

1.If the union set of d is a, then b may be an empty set b= that is, x has no solution, and m=0b is not an empty set, and m=1 and -1 are sufficient and unnecessary conditions respectively, which means that the value range of the given option should be the true subset of the problem range.

3.Draw the number line and find the points 2 and -3 Now we find x-2 + x+3 The solution set of a is empty, then the minimum value of the value on the left side of the inequality is less than a, and the minimum value of the sum of the distances from x to 2 and x to -3 is 5, so a is greater than 5

4.The inequality can be reduced to the union of the solution set of x -6 - x 0 and x 0 and the solution set of x -6 + x 0 and x 0.

The solution set is -3 x 3

where x should be a solution of -1 or 1 or no solution.

Sufficient and not necessary i.e., x>1 can be launched, but cannot be launched by x>1.

1)x≥2x-2+x+3=2x+1≤a

When a is an empty set, then a>5

1)x>0

x²-6≤x

02) x<0

x²-6≤-x

3≤x<0

3) x=0 constant holds.

It's too complicated, I'm just a mixed score, and I'm not good at high school math.

The minus half of your answer didn't make it to the last step, the question was not clear, the question asked for a range of 2a-b, not the answer of 2(a-b)+b, as for why, you will understand when you learn the higher field of algebra.

It's simple.

A mistake that beginners make.

Codirectional inequalities can only be added but not subtracted.

Only those who are different can be subtracted.

The range of B is known.

You can find the range of -b.

Adding it up is like subtracting you.

Don't make that mistake in the future.

Looking forward to the best and good reviews!

Tongxiang should know what it means.

It's either greater than or all are less than hehe.

Your solution ignores the correlation between a and b, i.e., aa, but a you have expanded the scope of b, so you are wrong with the last +b, because your last +b invisibly puts a< b, ignored, but +a, the range of b is only in the previous qualification, and is still between a and the positive dichotomy, and in the end it must be +a, not b< p>

minus 2 minus 2 < 2 (a-b) < 0

When these two equations are added together, when 2(a-b) is close to minus 2, b is actually close to half of the square.

A is added to the correct solution, and a is close to the minus half.

That is, two additive inequalities can't be close to the boundary at the same time, so your method expands the range.

x2 or x2 categorical discussions.

Or move 1 to the left to pass the score.

∵a=，cr

a=, again x2

ax-a-1 0 can be changed to (x-a-1)(x+1) 0 when a+1=-1, (x-a-1)(x+1) 0 is (x+1)20, x=-1 can be obtained, and a=-2 satisfies the topic.

When a+1 -1, i.e., a -2, the solution of (x-a-1)(x+1) 0 satisfies -1 x a+1, there must be a+1 0, and the solution is a -1, and the value range of the real number a is (-2, -1).

When a+1 -1 is a -2, the solution of (x-a-1)(x+1) 0 satisfies a+1 x -1, there must be a+1 -2, and the solution is a -3, and the value range of the real number a is (-3, -2).

In summary, the value range of the real number a is (-3, -1).

x1+x2=-(-a)=a

x1*x2=-a-1

z is the set of integers, and there is only one integer in the complement set of the solution of the inequality from the meaning of the problem, then 0<=x1-x2<=2

Are you sure the one in your question isn't +ax?

The set of solutions that meet the above conditions is x>t, t [-1,0).

To get the solution set for the inequality, only the quadratic term can be 0. In this case, k = 3 or k = -1.

When k=-1, the inequality 1>0 is constant, and the solution set is the set of real numbers r, so the condition is not satisfied.

When k=3, the solution set of the inequality is x>-1 4, which satisfies this condition.

Cross multiplication.

x-7）(x+2)＞0

Just go down and you're good to go.

It is known that f(x) = x-1 - x+1 ; the set of solutions of inequality f(x) 1; ②。If f(x)>a -3a is constant for x r, find the take of a.

Range of values. Untie:. When x -1, f(x)=-(x-1)+(x+1)=2;When -1 x 1, f(x)=-(x-1)-(x+1)=-2x;

When x -1, f(x)=-(x-1)+(x+1)=2;Hence -2 f(x) 2;The solution set of f(x) 1 is: {x - a -3a is constant for x r, so there must be a -3a<-2;i.e. a -3a + 2 = (a-1) (a-2) < 0, so 1

Because x 2 + y 2 "2xy, and x 2 + y 2 "2, so 2xy "2, then xy "1, so x + y + xy "2 + 1, that is, less than or equal to 3

Because: ab 2>ab

So: ab(b-1)>0

b(ab-a)>0

and a>ab, ab-a<0

So: b<0

1-b>0

and from a>ab: a(1-b)>0

So: a>0

So, from ab 2>a it can be obtained:

b^2>1

So: b<-1 (b>1 discarded).

In summary, the value range of b is b<-1