Try to explain (the third power of X + the square of 5x + 4x-1) - (- the square of - X - 3x + the square of 2x - 3) + (the square of 8-7x-6x + the third power of x) The value of is independent of the value of X

Try to explain (the third power of X + the square of 5x + 4x-1) - (- the square of - X - 3x + the square of 2x - 3) + (the square of 8-7x-6x + the third power of x) The value of is independent of the value of X

The square of + 2x in the second bracket should be + 2x cube primitive = x? + 5x? + 4x-1 + X? + 3x-2x? + 3 + 8-7x-6x? + X? = (x? - 2x? + X? + (5x? + X? - 6x? + (4x + 3x-7x) + (- 1 + 3 + 8) = 0 + 0 + 0 + 10 = 10

In a square + 1,7x + 8,9-2x = 6,7 + 8 = 15,6x-3x = 3x, 3x-2y = 1,3x-2 = 6, xsquare + 2x-1, X-Y = 2, x + 3 = 1 / 2,1 / x + 3 = 4, there are () unary first-order equations

There are five, namely:
9-2x=6
6x-3x=3x
3x-2=6
x+3=1/2
1/x+3=4

Given that the solution of equation 3 (2x-5) - A-4 = ax on X is suitable for both inequalities x-4 ≥ 0 and 4-x ≥ 0, find the value of A

solution
The inequality x-4 ≥ 0 leads to x > = 4
4-x ≥ 0 is X

(2008 Yangzhou) if there is and only one of the two univariate quadratic equations AX2 + 2x-5 = 0 on X is between 0 and 1 (excluding 0 and 1), then the value range of a is () A. a＜3 B. a＞3 C. a＜-3 D. a＞-3

According to the meaning of the title:
When x = 0, the function y = AX2 + 2x-5 = - 5;
When x = 1, the function y = a + 2-5 = A-3
In addition, there is only one of the two univariate quadratic equations AX2 + 2x-5 = 0 between 0 and 1 (excluding 0 and 1),
So when x = 1, the graph of the function must be above the x-axis,
So y = A-3 > 0,
A > 3
Therefore, B

Given that the solution of equation 3 (2x-5) - A-A = ax on X is suitable for both inequalities x-4 > = 0 and 4-x > = 0, find the value of A Is there any other answer,

∵｛x-4≥0,4-x≥0｝
The solution is 4 ≥ x ≤ 4
So x = 4
3（2x-5)-α-α=αx
6x-15-2α=αx
α（2+x）=6x-15
α=（6x-15）/（2+x）
∵x=4
∴α=（6×4-15）/（2+4）
=9/6
=3/2
[1522681117 for you]
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(2008 Yangzhou) if there is and only one of the two univariate quadratic equations AX2 + 2x-5 = 0 on X is between 0 and 1 (excluding 0 and 1), then the value range of a is () A. a＜3 B. a＞3 C. a＜-3 D. a＞-3

According to the meaning of the title:
When x = 0, the function y = AX2 + 2x-5 = - 5;
When x = 1, the function y = a + 2-5 = A-3
In addition, there is only one of the two univariate quadratic equations AX2 + 2x-5 = 0 between 0 and 1 (excluding 0 and 1),
So when x = 1, the graph of the function must be above the x-axis,
So y = A-3 > 0,
A > 3
Therefore, B

It is known that the maximum integer solution of the inequality X-1 in 2 and x-3 in 4 is the solution of the equation 2x AX = 4. Find the value of A

Solving inequality X-1 in 2 + 1 in 4 x-3

If the inequality system {2x + 31 / 2 (x-3), the integer interpretation of the equation about X 2X-4 = ax root, find the value of A

The solution of 2x + 31 / 2 (x-3) is - 3

Known inequality 2x-1-2x + 1

2x-1-2x + 1

If the integral solution of the inequality system 2x + 3 < 1, x > 1 (x-3) is the root of the equation 2X-4 = ax about X, then find the value of a?

2x+3＜1,
2x<-2
x<-1
x＞½（x-3）
2x>x-3
x>-3'
-So x = - 2
Substituting 2X-4 = ax
-4-4=-2a
A=4