1) Ta có
$\left( \dfrac{7}{11} \right)^{3x+2} = \left( \dfrac{11}{7} \right)^{x^2}$
$<-> \left( \dfrac{7}{11} \right)^{3x+2} = \left( \dfrac{7}{11} \right)^{-x^2}$
$<-> -x^2 = 3x + 2$
$<-> x^2 + 3x + 2 = 0$
Vậy $x = -1$ hoặc $x = -2$.
2) Ta có
$3^{x-2} . 5^{x-1} . 7^x = 245$
$<-> 3^{x-2} . 5^{x-1} . 7^x = 3^0.5.7^2$
Do đó $x - 2 = 0, x-1 = 1, x = 2$
Suy ra $x = 2$.