$(3+\sqrt{7})^{x^2+2x}+(3-\sqrt{7})^{x^2+2x}=2^{\frac{x^2+2x}{2}+4}$
$\Leftrightarrow (3+\sqrt{7})^{x^2+2x}+(3-\sqrt{7})^{x^2+2x}+2[(3+\sqrt{7})(3-\sqrt{7})]^{\frac{x^2+2x}{2}}=16.2^{\frac{x^2+2x}{2}}+2[(3+\sqrt{7})(3-\sqrt{7})]^{\frac{x^2+2x}{2}}$
$\Leftrightarrow (3+\sqrt{7}+3-\sqrt{7})^{\frac{x^2+2x}{2}}=16.2^{\frac{x^2+2x}{2}}+2.2^{\frac{x^2+2x}{2}}$
$\Leftrightarrow 6^{\frac{x^2+2x}{2}}=18.2^{\frac{x^2+2x}{2}}$
$\Leftrightarrow 2^{\frac{x^2+2x}{2}}.3^{\frac{x^2+2x}{2}}=3^2.$$2^{\frac{x^2+2x}{2}+1}$
$\Leftrightarrow 3^{\frac{x^2+2x}{2}-2}=2$
Nhầm, mod xóa hộ với