Ta có: \(f\left(x\right)=x-\dfrac{5}{3}\)
\(\Leftrightarrow f\left(x+2\right)=x+2-\dfrac{5}{3}\)
\(\Leftrightarrow f\left(x+2\right)=x+\dfrac{1}{3}\)
mà f(x+2)=ax+b
nên \(a=1\) và \(b=\dfrac{1}{3}\)
\(\Leftrightarrow a+b=1+\dfrac{1}{3}=\dfrac{4}{3}\)