47 / () = (). 7, fill in the appropriate number in brackets to make the equation hold. There are several different ways to fill in?

47 / () = (). 7, fill in the appropriate number in brackets to make the equation hold. There are several different ways to fill in?

47/( )=( ).7
Divisor times quotient = 47-7 = 40
40=1×40=2×20=4×10=5×8
Divisor greater than remainder
therefore
47/(40)=(1).7
47/(20)=(2).7
47/(10)=(4).7
47/(8)=(5).7
4 kinds

In the equation "1 = 1 (& nbsp;) + 9 (& nbsp;)", fill in a positive integer in each bracket so that their sum is the smallest, then the two numbers to be filled in are___ ．

Let two numbers be x and Y respectively, then 1x + 9y = 1, and their sum is x + y ∵ x + y = (1x + 9y) (x + y) = 10 + YX + 9xy ≥ 10 + 2yx · 9xy = 16. If and only if YX = 9xy, that is, y = 3x, x + y is the smallest and 1x + 9y = 1, so x = 4, y = 12, so the answer is: 4, 12

Fill in one number in parentheses to make the two numbers opposite, and the equation holds. What number to fill in?
3 × () - 2 × () = 15, what number to fill in the brackets? Remember to make two numbers opposite!
thank you!

Let this number be x, followed by - x, then 3x + 2x = 15, and x = 3
So fill in 3 and - 3

Put the numbers 1 ~ 9 in brackets to make the equation hold. Each number can only be used once. () () / () = () - () = () / ()

138/69 = 7 - 5 = 4/2
186/93 = 7 - 5 = 4/2
158/79 = 4 - 2 = 6/3
184/92 = 7 - 5 = 6/3
376/94 = 5 - 1 = 8/2

Fill in the seven numbers 1234567 in brackets to make the equation hold
()()÷()＝()×()＝()()
Each can only be used once

72*4=3*6==18

() 8 () × () = 70 () 5, what number in brackets makes the equation hold

785*9=7065

What are the numbers of a, B, C tables in integers 1 to 9 to make the above equation true?

c÷b=b÷a=a÷1
8÷4=4÷2=2÷1

Fill in the same number in () in the following equation to make the equation hold () * 9-4 * (2 * () + 3) = (() + 4) / 3

There is an equation
9X - 4 (2X + 3) = (X + 4)/3
The solution is x = 20
20×9 - 4×[2×20 + 3] = (20 + 4)/3

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

X-square + (a + b) x + AB + cx-10c = x-square-10x-11
(A+B+C+10)X+AB-10C+11=0
It's not about X
Then a + B + C + 10 = 0, ab-10c + 11 = 0

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