Ta có: \(F\left(1\right)=2\)
\(\Leftrightarrow a\cdot1+b=2\)
\(\Leftrightarrow a+b=2\)
hay a=2-b
Ta có: F(-2)=1
\(\Leftrightarrow a\cdot\left(-2\right)+b=1\)
\(\Leftrightarrow\left(2-b\right)\cdot\left(-2\right)+b=1\)
\(\Leftrightarrow-4+2b+b=1\)
\(\Leftrightarrow-4+3b=1\)
\(\Leftrightarrow3b-4=1\)
\(\Leftrightarrow3b=5\)
hay \(b=\frac{5}{3}\)
Ta có: F(1)=2
\(\Leftrightarrow a\cdot1+\frac{5}{3}=2\)
\(\Leftrightarrow a+\frac{5}{3}=2\)
hay \(a=2-\frac{5}{3}=\frac{1}{3}\)
Vậy: \(\left(a,b\right)=\left(\frac{1}{3};\frac{5}{3}\right)\)