Đáp án:1.$f(x)=-4x+3$
2.$f(x)=\dfrac{\left(4x-1\right)\left(x-1\right)}{2x^2-4x+3}$
Giải thích các bước giải:
1.Đặt $3a+1=x\to a=\dfrac{x-1}{3}$
Vì $f(3a+1)=-1-12a$
$\to f(x)=-1-12\cdot \dfrac{x-1}{3}$
$\to f(x)=-4x+3$
2.Đặt $\dfrac{a+1}{a}=x$
$\to 1+\dfrac1a=x$
$\to \dfrac1a=x-1$
$\to a=\dfrac{1}{x-1}$
Mà $f(\dfrac{a+1}{a})=\dfrac{3a+4}{a^2+2}$
$\to f(x)=\dfrac{3\cdot \dfrac{1}{x-1}+4}{(\dfrac{1}{x-1})^2+2}$
$\to f(x)=\dfrac{\left(4x-1\right)\left(x-1\right)}{2x^2-4x+3}$