Question

( 0,25)^2-x>(256/2)^x+3 решите неравенство

Answer 1

$(0,25)^{2-x}>(frac{256}{2})^{x+3}\ left(frac{1}{4}right)^{2-x}>128^{x+3}\ left(2^{-2}right)^{2-x}>left(2^7right)^{x+3}\ 2^{-2(2-x)}>2^{7(x+3)}\ 2^{-4+2x}>2^{7x+21}\ -4+2x>7x+21\ 5x<-25\ x<-5$