Докажи 16/f(x^2)=(g(2/x))^-1 Даны функции y=f(x) и g(x), где f(x)=x^2, g(x)=x^-4
f(x^2)=(x^2)^(–2/3)=x^(-4/3) 16/(f(x^2))=16/x^(-4/3)=16x^(4/3)=16x∛3 g(2/x)=(1/3*(2/x))=x/6 (g(2/x))^-(1)=(x/6)^(-1)=6/x 16x∛3 ≠ 6/x