When Wang Ming calculates the division with remainder, he takes the divisor 125 as 152. The quotient of the result is three times larger than the correct result, but the remainder is exactly the same

When Wang Ming calculates the division with remainder, he takes the divisor 125 as 152. The quotient of the result is three times larger than the correct result, but the remainder is exactly the same

So 152-125 = 27, 27 can be divided by divisor, and the result should be 3
27/3=9
The divisor is 9
125/9=13*9+8

In division without remainder, if the divisor + quotient x divisor = 798, what is the divisor?

Divisor △ divisor = quotient, divisor = divisor × quotient, then divisor × quotient is regarded as a share, and divisor is also regarded as a share. With 798 △ 1 + 1 × 1, we can find out the number of divisor. It is equal to 399

In division without remainder, the divisor + divisor x quotient = 84, what is the divisor

Divisor / divisor = quotient, then divisor = divisor x quotient, from the question get divisor + divisor x quotient = 2, divisor = 84, divisor = 42

_____ +The square of 3x - 5x + 2 = the square of 2x - 4x
Find the number on the horizontal line,

After moving the item, you can get what should be filled in the front
-X²+X-2

The known formula is s = VT + (1 / 2at & sup2;)
(1) If s, t, a are known, V is obtained;
(2) If s, V, t are known, then a

(1) From S = VT + (1 / 2at & sup2;)
That is, 2S = 2vt + at & sup2;
2vt=2s-at²
So v = (2S at & sup2;) / 2T
(2) From S = VT + (1 / 2at & sup2;)
That is, 2S = 2vt + at & sup2;
at²=2s-2vt
So a = (2s-2vt) / T & sup2;

Add brackets to the polynomial to get: A-B + C-D = a - (uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuy) ．

A-B + C-D = a - (B-C + D), so the answer is: B-C + D

Add brackets as follows
-3x to the third power + 4x to the second power - 2x to the third power, y to the second power - x + 2
(1) The coefficient of the highest order term becomes a positive number;
(2) The coefficient of quadratic term becomes positive;
(3) Put the odd items in the brackets with the "+" sign in front, and the rest items in the brackets with the "+" sign in front

-3x^3+4x^2-2x^3y^2-x+2
(1)-3x^3+4x^2+(-2y)^2x^3-x+2
(2)-3x^3+(4x)^2-2x^3y^2-x+2
(3)-(3x^3+2x^3y^2+x)+(4x^2+2)

The first volume of junior high school mathematics "using letters to represent numbers" is a chapter without brackets
I don't know how to simplify it, for example
If a ＞ 0, B ＜ 0, the reduction is | 5-b ︱ b-2a ︱ 1 ︱ 189; a (1 and half times a)
How to do this kind of topic!
Be sure to explain in detail, and then please tell me the solution of this type of problem in the future
Thank you!

Because B < 0, so 5-b must be greater than 5, that must be greater than 0; so its absolute value is itself, that is, 5-b
Because a ＞ 0, B ＜ 0, a negative number minus a positive number, wouldn't it be smaller

First volume of junior high school mathematics integral addition and subtraction brackets will be calculated, but how to add and subtract
For example: - x + (2x-2) - (3x + 5) I just don't know where to put the addition and subtraction
=-x+2x-2-3x-5

-X + (2x-2) - (3x + 5) I just don't know where to put the addition and subtraction
=-x+2x-2-3x-5
=(-1+2-3)x+(-2-5)
=-2x-7;
I'm very glad to answer your questions. Skyhunter 002 will answer your questions
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How to get rid of parentheses in mathematics integral in Volume 1 of junior high school?

The front is the plus sign, the front is the minus sign, the minus sign, the plus sign, the minus sign, the minus sign