When x takes what value, the fraction (x ^ 4 + x ^ 3-2) / (x ^ 3-x ^ 2 + x-1) * (x ^ 4-1) / (x ^ 2 + 2x ^ 2 + 2x + 2) / (x ^ 3-x-x ^ 2 + 1) / (- 2) When x takes what value, the value of fraction (x ^ 4 + x ^ 3-2) / (x ^ 3-x ^ 2 + x-1) * (x ^ 4-1) / (x ^ 2 + 2x ^ 2 + 2x + 2) / (x ^ 3-x-x ^ 2 + 1) / (- 2) can be positive integers

(x^4+x^3-2)/(x^3-x^2+x-1)*(x^4-1)/(x^3+2x^2+2x+2)÷(x^3-x-x^2+1)/(-2)
=[(x-1)(x^3+2x^2+2x+2)/(x-1)(x^2+1)]*[(x^2+1)(x+1)(x-1)/(x^3+2x^2+2x+2)]*[-2/(x-1)^2(x+1)]
=-2(x-1)(x^3+2x^2+2x+2)(x^2+1)(x+1)(x-1)/[(x-1)(x^2+1)(x^3+2x^2+2x+2)(x-1)^2(x+1)]
=-2 / (x-1) is a positive integer
So X-1 = - 1, X-1 = - 2
x=0,x=-1

If X-2 of fraction 2x + 3 is the range of positive X If the fraction 2x + 3 / x-21, the value of 0.2 is positive, the value of 3 is negative, and the value of 4 is 1 The range of X

1. The value is 0
That is, X-2 = 0 and 2x + 3 ≠ 0, x = 2
2. The value is positive
That is, X-2 > 0, 2x + 3 > 0, or X-2 < 0, 2x + 3 < 0, the solution x > 2 or x < 1.5
3. The value is negative
That is, X-2 > 0, 2x + 3 < 0, or X-2 < 0, 2x + 3 > 0, the solution is - 1.5 < x < 2
4. The value is 1
That is, X-2 = 2x + 3 and 2x + 3 ≠ 0, x = - 5

Calculate ((a) / (a-b) - (a) / (a + b)) / (2b) / (a? - B?)

The original formula = a [(a + B-A + b) / (a + b) (a-b)] × [(a + b) 9a-b) / 2B]
=2ab/2b
=a

47026=4x()+7x()+2x()+6x()

47026=4x(10000)+7x(1000)+2x(10)+6x(1)

Given 3x2-x-1 = 0, find the value of 6x3 + 7x2-5x + 1999

∵3x2-x-1=0
∴3x2-x=1
∴6x3+7x2-5x+1999
=2x（3x2-x）+9x2-5x+1999
=9x2-3x+1999
=3（3x2-x）+1999
=3+1999
=2002

6X ^ 2Y + 2xy-3x ^ 2Y ^ 2-7x-5yx-4y ^ 2x ^ 2-6x ^ 2Y, where | x + 3 | + (Y-2) ^ 2 = 0 ^2 is the square,

6x^2y+2xy-3x^2y^2-7x-5yx-4y^2x^2-6x^2y,
=-7x^2y^2-3xy-7x
Where | x + 3 | + (Y-2) ^ 2 = 0
That is, x + 3 = 0, Y-2 = 0
x=-3,y=2
The original formula = - 7 * 9 * 4-3 (- 3) * 2-7 (- 3) = - 252 + 18 + 21 = - 213

Given that the solution of equation (X-5) / 2-A + 1 = ax is suitable for inequalities - 1 / 2x ≥ - 1 and X-2 ≥ 0, we can find the value of A It takes a little process

-1/2x≥-1
x≤2
x-2≥0
x≥2
So, x = 2
By substituting (X-5) / 2-A + 1 = ax, we can get the following results
(2-5)/2-a+1=2a
-1/2=3a
a=-1/6

If there is and only one of the two quadratic equations ax ^ 2 + 2x-5 = 0 with respect to X is between 1 and 0 (excluding 0 and 1), then the value range of a is ()

There is only one between 1 and 0
Then the quadratic function ax ^ 2 + 2x-5 has only one intersection point with X continent between 1 and 0
So one of the function values of x = 0 and x = 1 is greater than 0 and the other is less than 0
So the multiplication is less than 0
x=0,ax^2+2x-5=-5
x=1,ax^2+2x-5=a+2-5=a-3
So - 5 (A-3) < 0
a-3>0
A>3

Only if the quadratic equation AX ^ 2 + 2x-5 = 0 has and only one root between 0 and 1 (excluding 0 and 1), then the value range of a is One and only one of the two roots is changed to have and only one becomes a

x=0,y<0
So when x = 1, Y > 0
A-3 > 0
A>3

If the two roots of the quadratic equation AX 2 + 2x-5 = 0 with respect to X are between 0 and 1 (excluding 0 and 1), then the value range of a is A3 A-3 Choose one of the four options above!

Exclusion method: for quadratic equation of one variable, then a is not equal to zero. A-3 exclusion; according to the combination of number and shape, the symmetry axis is a straight line x = - 2 / 2a, 0