\left \{ {{y=9-x} \atop {2^x-2^{9-x}=16}} \right.\\\
2^x-\frac{2^9}{2^x}=16\\\
2^{2x}-512=16*2^x\\\
2^{2x}-16*2^x-512=0\\\
2^x=t\\\
t^2-16t-512=0\\\
D=256+2048=2304=48^2\\\
t_1=\frac{16-48}{2}=-16\ \ \ \ \ t_2=\frac{16+48}{2}=32" alt=" \left \{ {{x+y=9} \atop {2^x-2^y=16}} \right. <=> \left \{ {{y=9-x} \atop {2^x-2^{9-x}=16}} \right.\\\
2^x-\frac{2^9}{2^x}=16\\\
2^{2x}-512=16*2^x\\\
2^{2x}-16*2^x-512=0\\\
2^x=t\\\
t^2-16t-512=0\\\
D=256+2048=2304=48^2\\\
t_1=\frac{16-48}{2}=-16\ \ \ \ \ t_2=\frac{16+48}{2}=32" align="absmiddle" class="latex-formula">
-16 не удовлетворяет области значения квадратичной функции
![2^x=32\\\
2^x=2^5\\\
x=5\\\
y=9-5=4](https://tex.z-dn.net/?f=2%5Ex%3D32%5C%5C%5C%0A2%5Ex%3D2%5E5%5C%5C%5C%0Ax%3D5%5C%5C%5C%0Ay%3D9-5%3D4%0A "2^x=32\\\
2^x=2^5\\\
x=5\\\
y=9-5=4")
Ответ: (5; 4)