Solution inequality in high school 5, solution inequality in high school

x^2-4x+3=(x-3)(x-1).

Mark two dots 1,3 on the number line, and draw a hollow dot because x 2-4x+3<0. Then draw a curve from the top right and go down from point 3 and up from point 1. Since the value taken is less than zero.

Therefore, the range of x is the value of the line below the number line. i.e. x (1,3).

If you don't want to do it, discuss it separately.

1.(x-3)>0,(x-1)<0

2.(x-3)<0,(x-1)>0

x^2-4x+3<0

Factoring yields (x-1)(x-3)<0

If (x-1)(x-3)<0 is true, then it should be used.

x-1>0 and x-3<0 or x-1<0 and x-3>0 The first inequality group solution set is 1, and the second inequality group solution set is an empty set.

Finally, go to the union of the two inequality solution sets to get the answer:

x^2-4x+3<0

3 is greater than 0x=7 2 or 1 2

Since the x 2-4x+3=y image is a downward protrusion, the value of a greater than 0x is 1 2, less than x and less than 7 2

First of all, you already know that =3 is greater than zero, so there are two unequal solutions to a quadratic equation, and you can find the two roots, and you can roughly draw the parabola on the number draw, because it requires a solution less than zero, so you can look at the part of the parabola where y is less than zero, and you can write the value, which is equivalent to being in the middle of the two roots.

with"Cross multiplication"Dharma. The solution = (x-3)(x-1)<0 gives x1=1, x2=3

It's better to be specific.

According to the inequalityDifferent typesto provide a solution.

It is mainly divided into basic inequality theorem, unary one-dimensional or quadratic inequality, inequality fraction, and parametric inequality.

UtilizedMathematical Thoughts:

1. The idea of categorical discussion; 2. The idea of combining numbers and shapes; 3. The idea of equalization and unequalization.

1.ForFundamental inequality theorem

Formula Law Shed Bizi:

2.ForUnary one-time inequality

3.ForUnary quadratic inequality

Solving process: Normalization: The coefficient before the quadratic term is positive.

Second judgment: Determine the root of the corresponding equation.

Three finds: find the root of the corresponding equation.

Four Drawings: Draw an image of the corresponding function.

Five-solution set: Write a solution set of inequalities based on images.

Note: When solving inequalities such as ax2+bx+c>0, the criteria for classification discussion are:

1. Discuss the size of a and 0; 2. Discuss the size of 0; 3. Discuss the size of the two roots;

The problem of the distribution of the roots of a quadratic equation:

Method: According to the image features of the quadratic function, from: opening direction, discriminant, axis of symmetry,

The group of inequalities is listed from the three perspectives of the function value, and in short, it is converted into a group of quadratic inequalities to solve.

4.ForFractional inequalities:

5.ForInequality with arguments:

1.Extract common factor 2Factorization 3Zoom in and out to deform 4Think of the argument as an unknown to exchange the principal element.

6.SolutionLinear programming problemsGeneral steps:

The first step is to make a feasible domain in a planar Cartesian coordinate system;

Step 2: Find the point corresponding to the optimal solution in the feasible domain;

Step 3: Solve the optimal solution of the equation to find the maximum or minimum value of the objective function.

7.YesInequalities of absolute valuesThere are two basic types of rubber chains.

Type 1: Make a a positive. According to the meaning of the absolute value, the absolute value of x is less than a, and its geometric meaning is that the set of points on the number line whose distance from the origin is less than a is the open interval (a, a).

Type 2: Let a be a positive number. According to the meaning of the absolute value, the solution set of the inequality is that the absolute value of x is greater than a, and its geometric meaning is that the set of points on the number line whose distance from the origin is greater than a is the union of two open intervals.

1).（x² -4x- 5 ) x² +x+ 1) (16-x²) 0

(x-2)²-9][(x+1/2)²+3/4](16-x²)≥0

Since (x+1 2) +3 4 is always greater than 0, it can be reduced to.

x-2)²-9](16-x²)≥0

i.e. [(x-2) -9] 0 and 16-x 0

or ( x-2) -9] 0 and 16-x 0).

solution -4 x -1; There is no solution, so the solution is -4 x -1

2), it should be (4x -9) (x +2x+3) 0 or something.

Original = (4x -9) [(x+1) +2] <0

The denominator is always greater than 0So the numerator must be less than 0, then there is:

4x²-9<0

>-3 2 -3 2x 4 or x -1

x²-3x＜18

x-3/2)²<81/4

=> -32 is 4 x 6, so the integer is solved.

If an inequality like this is squared, my suggestion is that it's better to make up (nx+k) +b greater than 0 or less than 0, and then it's easier to add the judgment. Hope it helps.

There are two cases: 1, x<0, 1 x<0<1 is naturally satisfied, and x<-1 is obtained according to -1<1 x2. When x>0, 1 x>-1 is naturally satisfied, and x>1 is obtained according to 1 x<1. To sum up, x<-1 or x>1

Divide into two inequalities, classify and discuss the relationship between x and 0, and solve the solution set of the two inequalities to find the union.

If x is greater than 0

x is greater than 1 if x is less than 0

x is less than 1

1) When the quadratic coefficient of the unary quadratic inequality with parameters is constant, but it is not known whether the corresponding quadratic equation of the beam chain has a solution to the slag number, the discriminant formula needs to be discussed. (2) When the quadratic coefficient of the unary quadratic inequality with parameters is constant, and the corresponding unary quadratic equation has two solutions, but the magnitude of the two solutions is not known, the size of the solution needs to be discussed. (3) When the quadratic coefficient of a quadratic inequality with parameters contains parameters, the quadratic coefficient should be discussed firstly, and secondly, the discriminant formula of the unary quadratic equation should be discussed, and sometimes the size of the solution of the equation should be discussed.

Steps to solve an inequality:

1. Remove the denominator;

2. Remove brackets;

3. Transferring items and merging similar items;

4. The coefficient is solved after it is reduced to one.

Precautions: 1. Multiply or divide a negative number on both sides of the inequality, and change the direction of the inequality sign;

2. If it is larger than both values, it will be larger than the big one, and if it is smaller than both values, it will be smaller than the small one; and leakage.

3. The two sides of the inequality are absolutely rotten to add or subtract each other, the same number or formula, the direction of the unequal sign remains unchanged, and the shift term should be changed;

4. Add or subtract both sides of the inequality, and the direction of the unequal sign remains unchanged for the same number or formula;

5. Multiply or divide by the same negative reflex on both sides of the inequality, and the direction of the inequality sign changes.

This question cannot be counted in this way, it should be divided into two steps.

From the question, we can know that x≠0

When x>0.

2x-1≥0

x again: (2x-1) x 3

2x-1≥3x

Solution: x -1

No solution. When x<0

2x-1≤0

Solution x again: (2x-1) x 3, and x<0

2x-1|/|x|≥3

2|x|+1≥3|x|

x|≥-1x|1x [-1 0)u(0 1], and x<0;

So the solution set for x is [-1 0).

Both sides divided by m become. -M compares with e, I want to discuss, annoying, don't do it!!

Solution: (x+4) *x+5) 2 * 2-x) 3 * x 2-x+1) *x 2-x-2)>0

Because (x+5) 2 0 then (x+5) 2≠0,x≠-5(x 2-x-2)=(x+1)(x-2).

x+4) *2-x) 3 * x 2-x+1) *x 2-x-2)=-(x+4) *2-x) 4 * x 2-x+1) *x+1)=-(x+4) *2-x) 4 * x 3+1)>0 (2-x) 4 0 then (2-x) 4≠0,x≠2-(x+4)* x 3+1)>0

x+4)* x^3+1）<0

Column inequality groups.

x+4>0

x^3+1<0

Solution: -40

Solution: x<-4, x>-1, no solution.

Composite Solution Set The solution set that yields the inequality is: -4

This is easy, using the zero-point interval method. Assuming that the equation inside each parenthesis is 0, it is marked on the number line in order and then underlined. The answer is -4

It can be resolved first.

x+4)*(x+5)^2*(2-x)^3*(x^2-x+1)*(x+1)*(x-2)>0

x+4)*(x+5) 2*(2-x) 4*(x 3+1)<0 gives (x+4)*(x 3+1)<0 and x≠-5 and 2 finally get -4

The fourth formula has no solution to Evergrande in 0, and it is directly ignored, and the last formula is decomposed into (x-2)(x+1), and the whole formula becomes (x+4) *x+5) 2 *(x-2) 4*(x+1)<0

Even times squared are equal to zero, so as long as they are not equal to -5,2, they can be omitted. So the solution set is -4

<> also has a little skill, one answer and two sentences, the hole is pure and can't be explained, and it's trembling and blind.