Question

4^×+4^3-×=20 помогите пж

Answer 1

$4^x+4^{3-x}=20\\4^x+\frac{4^3}{4^x} =20\\4^x=t,t\ \textgreater \ 0\\t+\frac{64}{t} =20\\t \neq 0\\t^2+64-20t\\t^2-20t+64=0\\D_1=100-64=36=6^2\\t_1=10+6=16\\t_2=4\\4^x=16\\x=2\\4^x=4\\x=1$