Given integers a, B, C, make the equation (x + a) (x + b) + C (X-10) = (x-1) (x + 1) hold for any x, find the value of C

Given integers a, B, C, make the equation (x + a) (x + b) + C (X-10) = (x-1) (x + 1) hold for any x, find the value of C

The title is wrong. Please check it
(X-11) (x + 1)
The solution is (x + a) (x + b) + C (X-10) = (X-11) (x + 1)
x^2+(a+b+c)x+ab-10c=x^2-10x-11
(a+b+c+10)x=10c-ab-11
Because the equation holds for any x, so
a+b+c+10=0
10c-ab-11=0
Namely
a+b=-10-c
ab=10c-11
a. B is the two integer solution of the equation x ^ 2 + (10 + C) x + (10-11) = 0
So} = (10 + C) ^ 2-4 (10c-11) is a complete square
100+20c+c^2-40c+44
=c^2-20c+144
144 = 12 ^ 2, C ^ 2-20c = 0 can satisfy the condition
C = 0 or C = 20

Given integers a, B and C, the equation (x + a) (x + b) + C (X-10) = (X-11) (x + 1) holds for any value of X. find the value of C

So a + B + C = - 10, and ab-10c = - 11, eliminating the parameter C, we get 10A + 10B + AB = - 111, that is, (a + 10) (B + 10) = - 11. Because a and B are integers, and - 11 = (- 1) × 11 = 1 × (- 11), a + 10 = 1, - 1, 11, - 11, B + 10 = - 11, 11, - 1, 1, so a + B + 20 = - 10, 10, that is, a + B = - 30, - 10 And C = 20 or 0

Given the integer ABC, the equation (x + a) (x + b) + C (X-10) = (x-1) (x + 1) holds for any X?

a+b+c=0
ab-10c=-1
ab+10(a+b)+1=0
That is, (a + 10) (B + 10) = 99 = 1 * 99 = 99 * 1 = 3 * 33 (corresponding to 3 pairs of negative numbers)
AB + 1 = 10C (because 99 decomposes 6 pairs, B of the last 3 pairs is the same as a of the first 3 pairs, a of the last 3 pairs is the same as B of the first 3 pairs, but C = (AB + 1) / 10) only half of the solution is complete
Solve a, B and get C
c=-80 -16 0 120 56 40
6

If positive integers a and B make the equation a + (a + b) (a + B − 1) 2 = 2009 hold, then a=______ ，b=______ ．

The original equation is transformed as: (a + b) 2 + (a-b) = 4018, let a ≥ b > 0, x = a + B, y = A-B, ∧ x ≥ y ≥ 0, ∧ the original equation is transformed into: x2 + y = 4018, ∧ x2 ≤ 4018 ≤ x2 + X, ∧ x = 63, ∧ y = 4018-632 = 49, ∧ a + B = 63, A-B = 49, a = 56, B = 7

If a positive integer a, B makes the equation a + (a + b) (a + B + 1) / 2 = 2009
So a =? B =?

A + (a + b) (a + B + 1) / 2 = 2009
2a/2+[(a+b)^2+(a+b)]/2=2009
2a+(a+b)^2+(a+b)=2009*2=4018
2a+(a+b)^2+(a+b)=(a+b)^2+2*(a+b)+1-1+(a-b)=(a+b+1)^2+(a-b)-1=4018
We get (a + B + 1) ^ 2 = 4018 + 1 - (a-b)
The squares closer to 4018 are 64 ^ 2 = 4096 and 63 ^ 2 = 3969
When a + B + 1 = 644018 + 1 - (a-b) = 4096, there is no solution
When a + B + 1 = 634018 + 1 - (a-b) = 3969
A = 56, B = 6

Find the value of positive integers a and B satisfying the equation a ^ 2 = B ^ 2 + 25

A ^ 2 = B ^ 2 + 25 A ^ 2-B ^ 2 = 25 (a + b) (a-b) = 25, because 25 can only be decomposed into 1 * 25 and 5 * 5. If a + B = 5, A-B = 5, then B = 0, not a positive integer. Abandon, so let a + B = 25, A-B = 1, and the solution is a = 13, B = 12

If positive integers a and B make the equation a + (a + b) (a + B − 1) 2 = 2009 hold, then a=______ ，b=______ ．

The original equation is transformed as: (a + b) 2 + (a-b) = 4018, let a ≥ b > 0, x = a + B, y = A-B, ∧ x ≥ y ≥ 0, ∧ the original equation is transformed into: x2 + y = 4018, ∧ x2 ≤ 4018 ≤ x2 + X, ∧ x = 63, ∧ y = 4018-632 = 49, ∧ a + B = 63, A-B = 49, a = 56, B = 7

In the equation = +, () fills in two different numbers to make the equation hold. How many different filling methods are there?
In the equation 1 / 10 = 1 / () + 1 / (), fill in two different numbers to make the equation hold. How many different filling methods are there?

Five methods were used
11 110
12 60
14 35
15 30
20 20
(when thinking, it's better to set the number of the previous bracket less than or equal to the number of the next bracket, so the number of the previous bracket less than or equal to 20 can be tested one by one.)

27÷______ =______… 3______ Fill in the appropriate number to make the equation true______ There are two different filling methods

Because quotient × divisor = 27-3 = 24; 24 = 2 × 2 × 2 × 3 = 23 × 3; (quotient, divisor) = (1,24), (2,12), (3,8), (4,6), (6,4), (8,3), (12,2), (24,1); from the fact that the remainder is smaller than the divisor, we can see that there are five ways to fill in divisor: 24,12,8,6,4

54/（ ）=（ ）…… 4. Fill in the appropriate number in the brackets to make the equation hold. How many different ways do you fill in?

54/（25 ）=（2 ）…… four
54/（5 ）=（10 ）…… four
54/（10 ）=（5 ）…… four