What is the solution to the equation (3/ m+/3-m)= (m^2+g/ m^2-g) m = 3 m = 6 all real numbers no solution

No solution.

Step-by-step explanation:

I must correct your given equation because it must read like this:

(3/(m+3)) - (m/(3-m)) = (m∧2+9)/(m∧2-9)  

Now we can solve it:

First we will transfer the fraction from the right to the left side of the equation.

(3/(m+3)) - (m/(3-m)) - (m∧2+9)/(m∧2-9) = 0

We will change binomial sign (3-m) = - (m-3) and that change sign of the second fraction from - to + , also we will change (m∧2-9) which is difference squared to (m-3)(m+3) and we get:

(3/(m+3)) + (m/(m-3)) - (m∧2+9)/(m-3)(m+3) = 0

The smallest common denominator for this equation is (m+3)(m-3)

when we multiply entire equation with (m+3)(m-3) we get:

3·(m-3) + m·(m+3)-(m∧2+9) = 0 => 3m-9+m∧2+3m-m∧2-9=0 =>

6m-18=0 => 6m=18 => m=18/6 = 3 => m=3

But we have conditions (m+3)(m-3) ≠ 0 => m+3≠0 and m-3≠0 => m≠-3 and m≠3

We conclude that this equation does not have solution.

Good luck!!!

i am thinking c. sorry if im wrong .

step-by-step explanation: