Xiao Ming is calculating the division with remainder. He mistakenly writes 308 as 368, and the quotient increases by 5, but the remainder is exactly the same. So he can find the divisor and remainder of the formula

Xiao Ming is calculating the division with remainder. He mistakenly writes 308 as 368, and the quotient increases by 5, but the remainder is exactly the same. So he can find the divisor and remainder of the formula

Divisor = (368 - 308) △ 5 = 12
308 ÷ 12 = 25 …… Yu 8
The divisor of the formula is 12 and the remainder is 8

First, when calculating the division with remainder, mischief mistakenly wrote the divisor 308 as 368. As a result, the quotient is increased by 5, and the remainder is exactly the same

Since the remainder is the same, only the quotient is increased by 5
So subtracting two numbers is a divisor of five
The divisor of 5 times is
368-308=60
So the divisor is 60 / 5 = 12
So the original division formula is
308 △ 12 = 25 + 8

The solution to the mathematical problem wrongly writes the divisor 308 as 368, and the result quotient increases by 5, while the remainder is exactly the same?

(368-308)/5=12
Divisor 12 and remainder 8

Given that the point m (4-2m, m-5) is on the bisector of the second and fourth quadrants, the coordinates of the point m are obtained

∵ point m (4-2m, m-5) is on the angular bisector of the second and fourth quadrants, and ∵ 4-2m + m-5 = 0. The solution is m = - 1, ∵ 4-2m = 4-2 × (- 1) = 4 + 2 = 6, m-5 = - 1-5 = - 6, ∵ point m (6, - 6)

In seventh grade mathematics, we know | a | = 8, | B | = 2, and | A-B | = B-A to find the value of a and B. We know | a | = 8, | B | = 2, and | A-B | = B-A to find the value of a and B

The absolute value of a number is equal to its opposite number, that is | n | = - N, then n ≤ 0, that is, n is a non positive number. Obviously, in the case of | A-B | = B-A, A-B ≤ 0, so a ≤ B. ∵ | a | = 8, | B | = 2, there are four combinations of a and B, that is, a B 828 - 2 - 8 - 2, in which there are two cases of a ≤ B: a = -

In the third year of junior high school mathematics, we know a / b = C / D = 3, and find the values of (a + b) / B and (c + D) d

a/b+1=3+1=4；
c/d+1=3+1=4

Given | a + 2 | + | B-3 | = 0, find the values of a and B

∵|a + 2 | + |b-3 | = 0, ∵ a + 2 = 0, B-3 = 0, the solution is a = - 2, B = 3

1-(2a-1)-(3a+3)
a-(5a-3b)+(2b+a)
-3(2s-5)+6s
3(-ab-2a)-(3a-b)
14(abc-2a)+3(6a-2abc)
3(xy-2z)+(-xy+3z)
-4(pq+pr)+(4pq+pr)

1-2a+1-3a+3=5-5a
a-5a+3b+2b+a=5b-3a
-6s+15+6s=15
-3ab-6a-3a+b=-3ab-9a+b
14abc-28a+18a-6abc=8abc-10a
3xy-6z+xy+3z=4xy-3z
-4pq-4pr+4pq+pr=-3pr

（1）(x+1)²+2(1-x)-x²
（2）（a+2)(a-2)-a(a+1)
（3）（a+3)²+a（2-a）
（4）（a+b)²+a(a-2b)

（1） 3
（2）-4-a
（3）8a+9
(4) 2a²+b²

12（x-2y）²-17（x-2y）³+13（2y-x）²-10（2y-x）³

=12（x-2y）²+13（x-2y）²-17（x-2y）³+10（x-2y）³
=25（x-2y）²-7（x-2y）³
=-18（x-2y）