Using monotonicity of function to prove inequality Prove x-x ^ 2 > 0, X ∈ (0,1) by monotonicity of function

Using monotonicity of function to prove inequality Prove x-x ^ 2 > 0, X ∈ (0,1) by monotonicity of function

The function increases monotonically at (0,1 / 2) and decreases monotonically at (1 / 2,1)
So there is a minimum when x = 0, or x = 1
f(x)>f(0)=0
That is, x-x ^ 2 > 0

By using the monotonicity of functions, the following inequality is proved: (1) x-x 2 ＞ 0, X ∈ (0,1)

If the square of X is less than x, it becomes two functions. Then, with the help of image analysis, monotonicity is analyzed

If x, y satisfy the inequality system x≤3 x+y≥0 X − y + 5 ≥ 0, can you draw the plane area where the point (x, y) is located?

As shown in the figure: draw straight line x = 3, x + y = 0, X-Y + 5 = 0 in rectangular coordinate system,
∵ the origin (0, 0) is not on the line X-Y + 5 = 0,
 by substituting the origin (0, 0) into X-Y + 5, we can see that the plane region where the origin is located represents the part of X-Y + 5 ≥ 0,
∵ the origin is on the line x + y = 0,
The plane region where the point (0, 1) is located represents the part of X + y ≥ 0

Draw the inequality group x + Y > 0, X= And the inequality (X-Y) (x-y-1)=

The first one is very simple, y = - X
Then Y > - x is above the line
X < = 2 means to the left of the line x = 2
The second takes X-Y as a variable
Then it means
0<(x-y)<1
Namely
x-1 It is sandwiched in the middle of two straight lines
See the figure below
Green indicates the area to be sought

The plane region represented by inequality (x + 2y-1) (X-Y + 3) > 0 is () A. B. C. D.

The inequality (x + 2y-1) (X-Y + 3) > 0 can be reduced to
x+2y−1＞0
X − y + 3 > 0, or
x+2y−1＜0
x−y+3＜0 ；
In the same coordinate plane, the plane region represented by these two inequalities is made, as shown in the figure;
From the graph, the plane area represented by the original inequality is C
Therefore, C

Draw the inequality group {X-Y + 5 > = 0, x + Y > = 0, X

It's very simple to draw this, but it's better not to look at the picture to calculate the whole number. It's easy to make mistakes, but it's fast and not easy to make mistakes
The first inequality plus the second inequality gives 2x + 5 > = 0, so - 2.5=

If x, y satisfy the inequality, if x, y satisfy the inequality group x ≤ 3, x + y ≥ 0, X-Y + 5 ≥ 0, please draw the plane area where the point (x, y) can be If x, y satisfy the inequality group x ≤ 3, x + y ≥ 0, X-Y + 5 ≥ 0, please tell us the plane region where the point (x, y) can be And how to draw x ≤ 3 in the number axis

The three inequalities are actually three straight lines: x = 3, y = - x, y = x + 5. Draw these three lines in a rectangular coordinate system, and then determine the upper, lower or left and right areas of the line according to the value of the inequality (for example, if x ≤ 3, take the left area of the line x = 3, including the point of the line itself)

It is known that there are three integer solutions of x-a ＞ 0 and 1-x ＞ 0, then the range of a is () Note "should: - 3"

By solving inequality (1), x > A is obtained
Solving inequality (2) leads to X

If the inequality system x-a > = 0 and 3-2x > - 1 has five integer solutions, then the value range of a is? It is found that a ≤ x < 2x has five integer solutions, which are 1, - 1, - 2, - 3 Why is the answer - 3 ≤ a ＜ - 4

∵x-a≥0,x≥a
3-2x＞-1,x＜2
∴a≤x＜2
∵ there are five integer solutions, which are 1,0, - 1, - 2, - 3
∴-3≤a＜-4

On the inequality system of X x−a＞0 If there are three integer solutions of 1 − x > 0, then the value range of a is______ .

From the inequality, x > A is obtained,
From inequality 2, x < 1 is obtained,
So the solution set of inequality system is a < x < 1,
∵ inequalities about X
x−a＞0
There are three integer solutions of 1 − x > 0,
The three integer solutions are 0, - 1, - 2,
The value range of a is - 3 ≤ a ＜ - 2