Three high school math questions about inequalities

1.With the matching method y=-2x +6x-m=-2(x -3x+9 4)+9 2-m=-2(x-3 2) +9 2-m, it is not difficult to see that if you want to make y constant less than 0, m should be greater than 9 2, as if you have typed the wrong answer!

2.|x-a|The solution set of <4 is -43 and the solution set is x-2>3 or x-2<-3, that is, x<-1 or x>5, if you want a b=r, then a-4<-1, and a+4>5, then the value range of a is: 1x+1-3, that is, -2>-2, which is impossible, this situation is not true.

If -1 x<2 then the inequality is: 2-x>x+1-3, then the range of x is: -1 x<2.

If x<-1, then the inequality can be reduced to: 2-x>-1-x-3, i.e., 2>-4, which is constant, so x<-1 also satisfies the requirement.

In summary, we can see that the set of solutions of inequalities is:

1 y=-2x +6x-m=-2(x -3x+9 4-9 4)-m=-2(x-3 2) +9 2-m If the first half is less than or equal to zero, the last two items 9 2-m are required to be less than zero, that is, m is greater than 9 2 Option A is eligible.

2 |x-a|<4 gets -43 and gets 35 or x<-1, so a+4>=5 and a-4<=-1, so 1<=a<=33 draw the graph of the inequality on both sides respectively, and compare it properly, when x is less than 2|x-2|In |x+1|-3 The above is therefore the solution set as.

The first course. If the value of y is always less than zero, then it means that its minimum value should also be less than zero, which should be understandable.

So, there are two ways to do it.

1.This analytically becomes a vertex (using the collocation method) to get the vertex (x,y), so the minimum value is y, so the value of y should be less than zero.

2,, find its axis of symmetry, bring its value in, and then get y, so the value of y should be less than zero.

The answer should be c

By the title, |x^2-4x+p|+|x-3|The solution of 5 is x3 on one side.

So, x=3 must be a critical point, if x>3, the left side will be "5", so, when x=3, |x^2-4x+p|+|x-3|=5 to get p=8 or -2

Substitute p=8 and p=-2 into |, respectivelyx^2-4x+p|+|x-3|5, p=-2 has to be discarded, so p=8.

Let [x]=a

4[x] 2-36[x]+45<0 to 4a 2-36a+45<0 to (2a-3)(2a-15)<0

That is, 3 2 and a = [x] is an integer solution to get a = 2, 3, 4, 5, 6, 7, so 2 x 8

The solution set is [2,8).

[x]=n

4[x] 2-36[x]+45<0 gives 4n 2-36n+45<0 and solves 3 2 when n=2 2<=x<3

When n=3, 3<=x<4

And so on.

1. The solution set of y=(a-2)x2+(a-2)x-4<0 is r, which means that the parabola y has no intersection point with the x-axis, and the opening is downward.

Then there is =(a-2) 2+16(a-2)<0, and a-2<0, the solution is -140, and when x 0, x 2-3|x|+2=x 2-3x+2=(x-1)(x-2)>0, the unequal plexcle solution is: x>1 and x>2 or 0 x<1 and 0 x<2 inequality is x>2 or 0 x<1

When x<0, x2-3|x|+2=x^2+3x+2=(x+1)(x+2)>0

The solution of the inequality is: 0>x>-1 and 0>x>-2 or x<-1 and x<-2 The solution of the inequality is -10 is: x>2 and ax>2 or x<2 and ax<2

When a>1, x>2>2 a, is demerged into x>2;x<2 a<2, and the solution is merged into x<2 a

The solution of the inequality is x>2 or x<2 a

When 02 a 2, the solution is merged into x>2 a; x<2 2 a, de-merge to x<2

The solution of the inequality is x>2 a or x<2

When a=0, the solution is empty and the solution is merged into x<2;

The solution of the inequality is x<2

When a<0, the solution is x<2 a<0, and the solution is empty; The x> segment accompanies 2 a, and the solution merges into 2 a, and the solution of the inequality is 2 a

5,ax 2-(2a+2)+4>0,b 2-4ac=(2a-2) 2 constant"=0

a=1 into a withered stand.

a>1 2/a>x,x>2

0x,x>2 Siege A

a《Yu Hui Nian 02 a

1.-2 2 or 0>x>-1, x<-2 5a = 1 x≠2

a>1 x> bridge repentance 2 or x<2 a a "leak elimination state 1 x>2 a or x<2

By rotation symmetry, you might as well set a b cThen there is 1 (bc) 1 (ac) 1 (ab), and 1 c 1 b 1 a, so a 12 (bc) + b 12 (ac) + c 12 (ab) is the ordinal sum, and since the sequential sum is not less than the out-of-order sum, there is.

A 12 (BC) +B 12 (AC) +C 12 (AB) A 12 (AC) +B 12 (AB) +C 12 (BC) = A 11 C + B 11 A + C 11 B (This is the out-of-order sum, which is not less than the inverse order).

a^11/a + b^11/b + c^11/c=a^10 + b^10 + c^10

The original inequality holds.

1.By the title, sina = 1 3, tanb = 3--- sinb = 3 2, cosc = 3 4--- sinc = 7 4

3/2=√12/4>√7/4---b>c√7/4=√63/12>√16/12=1/3---c>a∴b>c>a

2.From the title, there is: logb(1 b)0

by logb(1b)loga(b)<1

by loga(1b)loga(b)>0

a, b>1, then there are a>b>1

a, b<1, then there is 0b>1 or 01 c-a 4-1 4a-c 5

--1≤c≤7---2≤2c≤14---2≤8a-2c≤10

--6≤9a-3c≤9

--4≤9a-c≤23

--f(3)∈[4,23]

4.Less conditions, right?

Just state that under the condition of x 2 + y 2 + z 2 = 2.

The maximum value of the function f(x,y,z)=x+y+z-xyz should not exceed 2.

1) First of all, x 2 + y 2 + z 2 = 2 restricts the space point (x,y,z) on a sphere with a radius of root 2, and the function f(x,y,z) must have a maximum and minimum value on this sphere.

2) Sorry, the function f about x, y, z is only rotational symmetry, it is a bit troublesome to solve the minimum value point, you need to use calculus, and there are 8 sets of possible solutions in the end:

Groups 1 and 2: x = 0, y = z = 1 or -1

Groups 3 and 4: y = 0, x = z = 1 or -1

Group 5,6: z=0, x=y=1 or -1

Group 7,8: z=y=z=root(2 3) or - root(2:3).

3) Test the value of f on these 8 sets of solutions to the maximum point of 3 sets (f=2).

Group 1: x=0, y=z=1;Group 3: y=0, x=z=1;Group 5: z=0, x=y=1;

There are also 3 sets of minimum points (f=-2):

Group 2: x=0, y=z=-1;Group 4: y=0, x=z=-1;Group 6: z=0, x=y=-1;

The remaining 2 groups are saddle points (neither largest nor smallest).

This means that on a sphere -2<=f(x,y,z)<=2, or when x 2+y 2+z 2=2|x+y+z-xyz|<=2