Urgent, X squared + 3x + 2 / 2 X-1 + 2 + x-x squared - 4-x squared 10-x, OK fast The square of X + 3x + 2 parts of X-1 + 2 + 6 parts of x-x - 4 parts of X 10-x fraction operation

Urgent, X squared + 3x + 2 / 2 X-1 + 2 + x-x squared - 4-x squared 10-x, OK fast The square of X + 3x + 2 parts of X-1 + 2 + 6 parts of x-x - 4 parts of X 10-x fraction operation

(x's square + 3x + 2) of (x-1) + (2 + x-x's Square) of 6 - (4-x's Square) of (10-x) = (x + 1) (x + 2) of (x-1) + (2-x) (x + 1) of 6 - (2 + X) (2-x) of (10-x) = (x + 1) (x + 2) (2-x) of (x-1) (2-x) + (2-x) (x +

Simplify (x + 1) (x + 2) (x + 3) (x + 6) - 3x square
Factorization

（x+1）（x+2）（x+3）（x+6）-3x²
=（x+1）（x+6）（x+2）（x+3）-3x²
=（x²+7x+6）（x²+5x+6）-3x²
=（x²+6+6x+x）（x²+6+6x-x）-3x²
=（x²+6+6x）²-4x²
=（x²+6x+6-2x）（x²+6x+6+2x）
=（x²+4x+6）（x²+8x+6）

The first term of 3x squared - x = 7 is sharp

The first term of 3x squared - x = 7 is - X

If the square of x minus 3x = 4
Find the value of the fourth power of the algebraic formula 2x - the third power of 8x - the second power of 2x + 8x - 2007
Writing process

-2007
2x^4-8x^3-2x^2+8x-2007
=2x^3(x-4)-2x(x-4)-2007
=2x(x-4)(x^2-1)-2007
=2x(x-4)(x+1)(x-1)-2007
Given x ^ 2-3x = 4, (x-4) (x + 1) = 0
So the above formula = - 2007

How to calculate 3x square minus x minus 1

Let 3x ^ 2-x-1 = 0
Then x = 1 ± √ 13 / 6
So 3x ^ 2-x-1 = (6x-1 + √ 13) (6x-1 - √ 13) / 12

Square of X + 2 / 3x = 8 / 75

With the root formula: solution: X1 = 1 / 10, X2 = - 16

3x-5 / (xsquare-1) = A / (x-1) - B / (x + 1)
For example, find the value of integral A and B

(3x-5)/（x^2-1）=A/（x-1）-B/（x+1）
(3x-5)/(x+1)(x-1)=A(x+1)-B(x-1)/(x-1)(x+1)
So 3x-5 = ax + a-bx + B
So 3 = a-b
-5=A+B
So a = - 1, B = - 4

Given the square of a + B = 3x + X, the square of B + C + X, find the value of a-c

The square of 2x + X

If f (x) = MX ^ 2 - (M-4) x + 1 has at least one zero point to the left of the origin, find the value range of the real number M

(1) When m > 0 △ = (M-4) square - 4m = m square - 8m + 16 ≥ 0
So m ≥ (6 + 2 times root 5) or 0 < m ≤ 6-2 times root 5
When there are two zeros, but only one on the left, f (0) = 1

Given that f (x) = ax ^ 2 + (2a-3) x + A + 1 and at least one focus of X axis is on the right side of the origin, find the value range of real number a
Such as the title

Wrong!
Do not believe, with a 3 / 4 try. No real solution
1： When a = 0, f (x) = - 3x + 1, intersecting X-axis at (1 / 3,0), satisfying
2： When a ≠ 0,
Δ=(2a-3)^2-4a(a+1)=-16a+9≥0
A ≤ 9 / 16
When a = 9 / 16, the intersection x-axis of f satisfies at (5 / 3,0)
II when A0,
When a