The sum of the coefficients of the quadratic term, the coefficient of the first order term and the constant term of the quadratic equation of one variable 2x? - x = 6 is?

The sum of the coefficients of the quadratic term, the coefficient of the first order term and the constant term of the quadratic equation of one variable 2x? - x = 6 is?

2x²-x=6
That is 2x? X-6 = 0
That is, 2 + (- 1) + - 6 = - 5

2X + 3 / X (x-1) (x + 2) = A / B + B / X-1 + C / x + 2 (A.B.C is a constant), find the value of a.b.c

(2x+3)/[x(x-1)(x+2)]=A/x+B/(x-1)+C/(x+2)
=[A(x^2+x-2)+B(x^2+2x)+C(x^2-x)]/[x(x-1)(x+2)]
=[(A+B+C)x^2+(A+2B-C)x-2A]/[x(x-1)(x+2)]
The results are as follows:
A+B+C=0
A+2B-C=2
-2A=3
A=-3/2
B=5/3
C=-1/6

(x ^ 2 + 2x + 3) / ((x + 2) ^ 3) = A / (x + 2) + B / ((x + 2) ^ 2) + C / ((x + 2) ^ 3), find a, B, C (a, B, C are constants)

This is a case by case discussion
Firstly, the original equation is equivalent to (A-1) x ^ 2 + (4a + b-2) x + 4A + 2B + C-3 = 0 (x ≠ - 2)
Discuss in two cases
(1) When a ≠ 1, if Δ

Given (2x + 4) / [x (x-1) (x + 2)] = A / x + B / (x-1) + C (x + 2) (a, B, C are constants), find the values of a, B, C More steps! thank you!

By removing the denominator, we can get: 2x + 4 = a (x-1) (x + 2) + b * x (x + 2) + C * x (x-1) 2x + 4 = a (x? + X-2) + B (x? + 2x) + C (x? - x) 2x + 4 = (a + B + C) x? + (a + 2b-c) x-2a, and a + B + C = 0A + 2b-c = 2-2a = 4

If a is known x+1+B x−1＝2x+3 X2 − 1 (where a and B are constants), then a=______ ，B=______ .

The original equation can be changed into: a (x − 1) + B (x + 1)
x2−1=(A+B)x−A+B
x2−1=2x+3
x2−1，
So,
A+B＝2
- A + B = 3, a = - 0.5, B = 2.5

Given (2x + 3) / (x (x + 2) (x-1)) = (A / x) + (B / (x-1)) + (C / (x + 2)), calculate the values of constants a, B, C

(2x+3)÷[x(x-1)(x+2)]=A÷x+B÷(x-1)+C÷(x+2)
2x+3=A*(x-1)(x+2)+B*x(x+2)+C*x(x-1)
2x+3=A(x^2+x-2)+B(x^2+2x)+C(x^2-x)
2x+3=(A+B+C)x^2+(A+2B-C)x-2A
A+B+C=0
A+2B-C=2
-2A=3
By solving the equations, the following results are obtained
A=-3/2,B=5/3,C=-1/6

6X + 5 / 2 (2x + 2x + 1 / 2) = 10x + 3 / 4 to find the value of X

6x+5/2(2x+2x+1/2)=10x+3/4
6x+5X+5X+5/4=10x+3/4
16x-10x=3/4+5/4
6x=2
x=1/3

Given the square of X + 3x = 3, the square of - 2x - 6x + 5 Sorry, it is known that the square of X + 3x = 1, the square of - 2x - 6x + 5,

x²+3X=3
-2x²-6X+5=-2(x²+3x)+5
=-2*3+5
=-6+5
=-1

(the fourth power of x-6x cube-2x square + 18x + 23) / (x-3) (X-5), x = 19-8 times the root sign 3, and find the value of the above formula

19-8√3
=16-2√48+3
=(√16-√3)²
So x = 4 - √ 3
x-4=-√3
square
x²-8x+16=3
x²-8x=-13
Molecular = x ^ 4-8x? + 2x? - 2x? + 18x + 23
=x²(x²-8x)+2x³-2x²+18x+23
=-13x²+2x³-2x²+18x+23
=2x³-15x²+18x+23
=2x³-16x²+x²+18x+23
=2x(x²-8x)+x²+18x+23
=-26x+x²+18x+23
=x²-8x+23
=-13+23=10
Denominator = x? - 8x + 15
=-13+15
=2
Original formula = 5

Given the square of X + 3x = 1, find the square of - 2x - 6x + 5

solution
-2x²-6x+5
=-2(x²+3x)+5
=-2×1+5
=3