Solving a Fractional Inequality Mathematics Solving a Fractional Inequality Problem

>01-[x/(x^2+1)]^2>0

1+[x/(x^2+1)]}0

x 2+x+1) (x 2+1)]*x 2-x+1) (x 2+1)]>0 is very true.

x is taken as an effective number.

2.In contrast, x has no solution.

1.(1-x 2) (x 2+1) 2>0x 2+1) 2>0 Heng established.

So (1-x 2) >0

x^2<1

10 Heng was established.

So (1-x 2) >0

x^2>1

x>1 or x<-1

can be done with the commutation method.

The first pass: let x 2=t, t>=0

The original inequality is equivalent to 1-t (t+1) 2>0

After the general fraction, we can get: (t 2 + t + 1) (t 2 + 2t + 1) > 0 denominator is equal to 0, and the numerator is always equal to 0 (the formula is known), so the solution set of inequality is r.

The sign of the second path is completely opposite to the first, and the solution set is an empty set.

1-x^2/(x^2+1)^2>0

Because (x 2+1) 2>0, the two sides are directly multiplied by (x 2+1) 2x 4+2x 2+1-x 2>0

x^4+x^2+1>0

<0 is found, and the opening is up, so x r

In the same way, the second one is <0 and the opening is downward, so there is no solution.

Divide generally, shift terms, divide into integers, and solve integer inequalities.

Answer: x<-2 or x>= -5 3

3-1/(x+2)≥0

3x+6-1)/(x+2)≥0

3x+5)/(x+2)≥0

i.e. (3x+5)(x+2) 0

x≤-2，x≥-5/3

The denominator is not equal to 0

So x -2, x -5 3

1 – 3 Move 3 to the left i.e. (-3x-5) (x+2)<0

The solution of x+2 inequality is equivalent to: (-3x-5)(x+2)<0

So the solution is: x<-2 or x>-5 3

3-1/(x+2)≤0

3x+6-1)/(x+2)≤0

3x+5)/(x+2)≤<0

i.e. (3x+5)(x+2) 0

2≤x≤-5/3

The denominator is not equal to 0

So -2

1 x+2 is less than or equal to 3

1 is less than or equal to 3 (x+2).

1 less than or equal to 3x+6

5 Less than or equal to 3 x

x is greater than or equal to -5 3

1/(x+2)<=3

When x+2<0, yes, you get x<-2

When x+2>0, x+2>=1 3 gives x+2>=-5 3, so x<-2 or x>=-5 3

By the title, x is not equal to 1

If x>1, then x-1>0

Both sides of the inequality are multiplied by x-1 at the same time, and the sign remains the same.

x-5<=2(x-1)

x-5<=2x-2

Merge items of the same kind.

x>=-3

Then there is x>1 according to the assumption

If x<1

Both sides of the inequality are multiplied by x-1 at the same time, and the sign changes.

x-5>=2(x-1)

x-5>=2x-2

Merge items of the same kind.

x<=-3

Sum up. x<=-3 or x>1

By the title, x is not equal to 1

If x>1, then x-1>0, (x-5) (x-1)<=2 inequality is multiplied by x-1 on both sides at the same time, and the sign remains unchanged.

x-5)<=2(x-1)

x-5<=2x-2

5+2<=2x-x

3<=x

i.e. x>=-3

If x<1

Both sides of the inequality are multiplied by x-1 at the same time, and the sign changes.

x-5)>=2(x-1)

x-5>=2x-2

5+2<=2x-x

3>=x

i.e. x>=-3

So x<=-3 or x>1

(x-5)/(x-1)≤2

Move the item, got. x-5)/(x-1)-2≤0

x-5)-2(x-1)]/(x-1)≤0-(x+3)/(x-1)≤0

x+3)/(x-1)≥0

It can be solved by the number line stubbing method.

x<=-3 or x>1

and x-1 is the denominator.

x≠1x -3 or x>1

Solution: (x-5) (x-1) 2

x-5)/(x-1)-2≤0

x-5)/(x-1)-2(x-1)/(x-1)≤0(x+3)/(x-1)≥0

Its equivalent. x+3)(x-1) 0, and x-1≠0

So, x>1 or x-3

(1)x/(x-2)-a>0

x-ax+2a)/(x-2)>0

x-ax+2a)(x-2)>0

When (-2a(1-a))=2 there is no solution.

When (-2a(1-a))>2 22 (-2a(1-a))(2)x<-1 or 2=

（x-3）/（x^2+5x+4）≤ 0

From the question, we can get: x +5x+4≠0, that is, x≠ -4 and x≠ -1 are multiplied by x +5x+4.

x-3 0 and x +5x+4>0 or x-3>=0 and x +5x+4<0

Because x +5x+4=(x+1)*(x+4) then x<=3 and (x<-4 or x>-1) or x>=3 and (-4<=x<=-1).

So x<-4 or -1

x (-1,3) and x (-4).

Solution: (x-3) (x 2+5x+4) 0

x-3)/[(x+1)(x+4)]≤0

It can be seen that there are three factors, sorted by size, as follows:

x-3、x+1、x+4

Because: the result of multiplication and division of these three factors is non-positive, so all three factors are either non-positive or have only one non-positive value.

And because x+1 and x+4 are in the denominator, x+1 and x+4 must not be 0, therefore, there are: x-3 0, x+1 0, x+4 0 .........1) Or:

x-3≤0、x+1＜0、x+4＜0………2) Solved from (1): x 3, x -1, x -4

Therefore, there is: -1 x 3

From (2), we get: x 3, x -1, x -4

Therefore, there is: x -4

To sum up, the solutions to the given inequalities are: x (-1,3) and x (-4).

（x-3）/（x^2+5x+4）≤ 0

From the question, we can get: x +5x+4≠0, that is, x≠ -4 and x≠ -1 are multiplied by x +5x+4.

x-3 0x3 so the solution set of inequalities is.

x 3 and x≠ -4 and x≠ -1

From the question, we can get: x +5x+4≠0, that is, x≠ -4 and x≠ -1 if x-3 0, then x +5x+4 0, i.e. x 3, (x+1)(x+4) 0, if x+1 0, x+4 0, then -1 x 3, if x+1 0, x+4 0, then x -4

If x-3 0, then x +5x+4 0, i.e. x 3,(x+1)(x+4) 0, then if x+1 0, x+4 0, then there is no solution.

If x+1 0, x+4 0, there is no solution.

So the solution is -1 x 3, x -4, not only between -1 x -4, you can bring a number to verify, for example, assuming x=2, is satisfied to be less than 0, assuming x=-5 is also less than 0.

x< (1 x) (1 x)<1 should pay attention to the positive and negative of x.

The first one: move the formula on the right to the left: x-1 x<0:

Score: (x 2-1) x <0

When x < 0, multiply x on both sides, change the sign, get: x 2-1>0, solve x>1 or x<-1, combine x<0, get x<-1

In the same way, when x>0, 01

x 2-1) is read as the square of x minus one.

For example, x n-1 is read as x to the nth power minus one.

Does it mean: 1 x >1?

In fact, it is not necessary to divide it, when x > 0, multiply x on both sides to get: 1>x, that is: 1>x>0

When x < 0, multiply x on both sides to get: 1 in total, 0

Assuming that fractional inequalities are written in the form of a b+c d e f (all polynomials with x are capitalized below, which may of course be constants), the following discussion is purely theoretical, and an example is given at the end.

Scores. It is not the same as the solution method of fractional equations, and the denominator cannot be removed at the beginning, because after multiplying the denominator at the same time, I don't know whether the unequal sign will change direction. All the denominators are made to the same, and the inequality becomes a'/r+c'/r≥e'r, r is the common denominator.

Move to simplification. Move the right side over and change the chain to (a'+c'-e'r 0, above a'+c'-e'It is possible to merge similar terms and reduce them into a single formula pIt eventually becomes p r 0

Decompose the factor. p and r are factored separately (generally speaking, it is difficult to decompose factors, but the problems of fractional inequalities are either not decomposed, or they are easy to decompose, and generally there are no problems that can be decomposed but are difficult to decompose), and then the numerator and denominator can be reduced to (p1p2....).pm)/(r1r2…rn) 0.

Convert to integer inequality. This step is crucial. We know that a b 0 and a b 0 are the same thing, because multiplication and division are the same for positive and negative signs.

Therefore (p1p2....pm)/(r1r2…rn) 0 is equivalent to.

p1×p2×…×pm)×(r1×r2×…rn) 0 and then the same solution as integer inequalities. However, it should be noted that there is a difference between fractional inequality and integer unequal nucleus Husun, and after the solution, it must be checked that the r1 rn that was originally used as the denominator is not 0, and cannot be marked with an equal sign (of course, the "" sign or -1 or x -5

I should have written this in great detail......But after all, I am not a teacher, so many of the languages are organized by myself, which may deviate from the authoritative textbooks of secondary schools or what teachers say. There are inevitably mistakes in it, and it is for reference only.