The square of 5x = 7x

The square of 5x = 7x

5x²=7x
5x²-7x=0
x(5x-7)=0
X = 0 or x = 7 / 5

If a / b = 21.7, and the sum of divisor, divisor, quotient and remainder is 541, then what are the divisor and divisor B

A = 21 * B + 7, a + B + 21 + 7 = 541, a = 490, B = 23

Is there a real number a such that the set of integer solutions of the inequality system 2x & sup2; + (2a + 5) x + 5A < 0; X & sup2; - X-2 > 0; is a single element set?

The solution. X & sup2; - X-2 = (X-2) (x + 1) > 0 gives x > 2 or X

The quotient of a divided by B is 21, and the remainder is 2!

Business is 21
The remainder is 2x10 = 20
If you don't understand this question, you can ask,

It is known that the solution set of real number a satisfying | 5A -- 3 | a ^ (2x ^ 2 + X -- 10) is (please write the procedure)

From | 5A -- 3|

The divisor divided by the divisor, the quotient is 21, the remainder is 17, the divisor is larger than 400, smaller than 500, and is a prime number
This is the problem in my exercise book. It's due soon,
You are too simple on the first floor.

437 400 / 21 = 19 ··· 1 500 / 21 = 23 ··· 17, so the divisor should be one of 19, 20, 21, 22, multiply by 21 and add 17 respectively, and finally get 479. At this time, the divisor is 22
Hehe, 437 is not prime number, so I'd like to change it!

It is known that the solution set of inequality (2a-b) x + a-5b > 0 is x < 0
Given that the solution set of inequality (2a-b) x + a-5b > 0 about X is x < 10 / 7, find the inequality ax-b > 0 > about X
Given that the solution set of inequality (2a-b) x + a-5b > 0 about X is x < 10 / 7, find the inequality ax-b > 0 about X

(2a-b)x>5b-a
Two sides divided by 2a-b is X

The quotient of a divided by B is 25, and the remainder is 21. When the divisor is the smallest, what is the divisor

The divisor should be greater than the remainder, and the minimum is 21 + 1 = 22
The divisor is:
22×25+21=571

Given that y = loga (2-ax) is an increasing function on [0,1], the solution set of the inequality loga | x + 1 | > loga | x-3 | is ()
A. {x | x ＜ - 1} B. {x | x ＜ 1} C. {x | x ＜ 1 and X ≠ - 1} D. {x | x ＞ 1}

So from the inequality loga | x + 1 | loga | x-3 | we can get 0 | x + 1 | x-3 |. X + 1 ≠ 0 (x − 3) 2 ＞ (x + 1) 2, we can get x | 1, and X ≠ - 1, so the solution set of the inequality is {x | x < 1, and X ≠ - 1}, so we choose C

The quotient of a divided by B is 5, and the remainder is 3. If the divisor and divisor are reduced by 10 times at the same time, the quotient is (), and the remainder is ()
Come on!

The quotient of a divided by B is 5 and the remainder is 3. If the divisor and divisor are reduced by 10 times at the same time, the quotient is (5) and the remainder is (0.3)