1. **State the problem:** Solve the equation $$\frac{x^2}{25}+\frac{(mx+5m)x^2}{9}=1$$ for $x$ in terms of $m$. 2. **Rewrite the equation:** $$\frac{x^2}{25} + \frac{(mx + 5m)x^2}{9} = 1$$ 3. **Distribute $x^2$ inside the second fraction:** $$\frac{x^2}{25} + \frac{m x^3 + 5 m x^2}{9} = 1$$ 4. **Multiply both sides by the common denominator $225$ (LCM of 25 and 9) to clear fractions:** $$225 \times \left(\frac{x^2}{25} + \frac{m x^3 + 5 m x^2}{9}\right) = 225 \times 1$$ 5. **Simplify each term:** $$9 x^2 + 25 (m x^3 + 5 m x^2) = 225$$ 6. **Distribute 25 inside the parentheses:** $$9 x^2 + 25 m x^3 + 125 m x^2 = 225$$ 7. **Group like terms:** $$25 m x^3 + (9 + 125 m) x^2 = 225$$ 8. **Rewrite as a cubic equation:** $$25 m x^3 + (9 + 125 m) x^2 - 225 = 0$$ 9. **This is a cubic equation in $x$ with coefficients depending on $m$.** 10. **Summary:** The solution for $x$ satisfies $$25 m x^3 + (9 + 125 m) x^2 - 225 = 0$$ This cubic can be solved using standard methods (Cardano's formula) or numerically for specific values of $m$. **Final answer:** $$25 m x^3 + (9 + 125 m) x^2 - 225 = 0$$