When m is an integer, the solution of the equation 3mx-10 = 3 (x-4 / 3) is a positive integer

When m is an integer, the solution of the equation 3mx-10 = 3 (x-4 / 3) is a positive integer

3mx-10=3x-4
3mx-3x=6
x(m-1)=2
m> 1, the equation is a positive integer solution

When the integer m=_____ The equation (M + 2) x + 2mx-4 has positive integer solution

The title should be as follows: (M + 2) x = 2mx-4
It is concluded that: (m-2) x = 4, x = 4 / (m-2)
To make x a positive integer, m must also be an integer
M-2 is the positive divisor of 4,
That is, m-2 = 1 or 2 or 4
M = 3 or 4 or 6
It can be adopted. It must be correct!

When m takes what value, the solution of the equation 1 / 2mx-5 / 3 = 1 / 2 (x-4 / 3) about X is a positive integer? (/ represents the fractional line)

Hello!
The beauty of Mathematics
1/2 mx - 5/3 = 1/2 (x - 4/3)
mx - 10/3 = x - 4/3
(m - 1)x = 2
x = 2/(m - 1)
If x is a positive integer, then M = 2 or 3

When m is an integer, the solution of the equation 12mx − 53 = 12 (x − 43) is a positive integer?

If we solve the equation 12mx − 53 = 12 (x − 43), remove the denominator, we get 3mx-10 = 3 (x-43), remove the bracket, we get 3mx-10 = 3x-4, transfer the term and merge the similar terms, we get x (m-1) = 2, the coefficient is 1, we get x = 2m − 1, ∵ the solution of the equation is a positive integer, ∵ x > 0 and is a positive integer, ∵ 2m − 1 > 0 and M is a positive integer, ∵ M-1 is a positive divisor of 2, that is, M-1 = 1 or 2, ∵ M = 2 or 3

3.14 * 20 * 20 △ 2-40x △ 2 = 28 * is the multiplier sign and X is the unknown of the equation

3.14*20*20÷2-40X÷2=28
628-20x=28
20x=628-28=600
x=600÷20
x=30

The solution of the equation PX + 8q = 333 of X is 1, and P, q are prime numbers, P

In group 1, P = 5, q = 41, P = 37, q = 37, P = 0

The root ratio of x ^ 2 + PX + q = 0 is 1:2, and the discriminant value is 1. Find the value of PQ and solve the equation

p² -4q=1
x1 +x2 =-p
x1*x2=q
x1：x2 =1：2
The solution is p = 3, q = 2; X1 = - 1, X2 = - 2
Or P = - 3, q = 2; X1 = 1, X2 = 2
So, PQ = 6 or - 6

On the range of y = (x2-x + 1) / (2x2-2x + 3)
First, the function is transformed as: (2y-1) X2 - (2y-1) x + (3y-1) = 0
If 2y-1 = 0, that is, y = 1 / 2, the left side is not equal to 0, so y is not equal to 1 / 2
Then, because x belongs to R, so △ = (2y-1) 2-4 (2y-1) (3y-1)) = 0
The question is, why can't we reduce (2y-1) at the same time and extract and solve it again? Isn't y equal to 1 / 2

Finding the range of the function is the condition for the equation to have a solution
Since x belongs to the real number set R, the equation must have a solution
As a parameter of the equation, y has influence on the solution of the equation
When 2y-1 = 0, the equation is a linear equation, which should be considered from the linear equation of one variable;
When 2y-1 ≠ 0, the equation is a quadratic equation with one variable
The main reason why 2y-1 can't be reduced is that the formula is an inequality, in which the reduction will change the direction of inequality. Now we don't know its positive and negative, so we can't reduce it

Y = 3 + 2cos (2x - π / 3) range X ∈ R

Because - 1 ≤ cos (2x - π / 3) ≤ 1
-2≤2cos(2x-π/3)≤2
1≤3+2cos(2x-π/3)≤5
So the range is [1,5]

Find the range of y = 2cos ^ x + 2sinx-4 in [- 2 π / 3,2 π / 3]

x∈[-2π/3,2π/3]
sinx∈【-1,1】
Let SiNx = t, t ∈ [- 1,1]
y=2(1-t²)+2t-4
=-2t²+2t-2
=-2(t-1/2)²-3/2
T = 1 / 2, the maximum is - 3 / 2
T = - 1, the minimum is - 6
Range [- 6, - 3 / 2]