Решить систему уравнений. 10 класс
век буду

ответ:

{(1; 1)}

пошаговое объяснение:

[tex]\left \{ {{9^{x}-5^{2y} +16=0} \atop {{9^{\frac{x}{2} }-5^{y} +2=0}} \right. \{ {{(9^{\frac{x}{2} }-5^{y})*(9^{\frac{x}{2} }+5^{y} ) +16=0} \atop {{9^{\frac{x}{2} }-5^{y} = -2}} \{ {{-2*(9^{\frac{x}{2} }+5^{y} ) = -16} \atop {{9^{\frac{x}{2} }-5^{y} = -2}} /tex]

[tex]\left \{ {{9^{\frac{x}{2} }+5^{y} = 8} \atop {{9^{\frac{x}{2} }-5^{y} = -2}} \{ {{2*9^{\frac{x}{2} }+5^{y}-5^{y} = 8+(-2)} \atop {{9^{\frac{x}{2} }-5^{y} = -2}} \{ {{2*9^{\frac{x}{2} } = 6} \atop {{9^{\frac{x}{2} }-5^{y} = -2}} \{ {{9^{\frac{x}{2} } = 3} \atop {{9^{\frac{x}{2} }-5^{y} = -2}} /tex]

[tex]\left \{ {{3^{2*\frac{x}{2} } = 3^{1}} \atop {{3^{2*\frac{x}{2} }-5^{y} = -2}} \{ {{x=1} \atop {{3^{x}-5^{y} = -2}} /tex]

[tex]\left \{ {{x=1} \atop {{3^{1}-5^{y} = -2}} \{ {{x=1} \atop {{5^{y} = 5^{1}}} \{ {{x=1} \atop {{y = 1}} \right.[/tex]