\(f\left(x\right)=ax-3\)
a) \(f\left(3\right)=a.3-3=9\)
\(\Leftrightarrow f\left(3\right)=a.3=9+3\)
\(\Leftrightarrow f\left(3\right)=a.3=12\)
\(\Leftrightarrow f\left(3\right)=a=12:3\)
\(\Leftrightarrow f\left(3\right)=a=4\)
Vậy \(f\left(3\right)=9\) thì \(a=4\)
b) \(f\left(5\right)=a.5-3=11\)
\(\Leftrightarrow f\left(5\right)=a.5=11+3\)
\(\Leftrightarrow f\left(5\right)=a.5=14\)
\(\Leftrightarrow f\left(5\right)=a=14:5\)
\(\Leftrightarrow f\left(5\right)=a=\dfrac{14}{5}\)
Vậy \(f\left(5\right)=11\) thì \(a=\dfrac{14}{5}\)
c) \(f\left(-1\right)=a.\left(-1\right)-3=6\)
\(\Leftrightarrow f\left(-1\right)=a.\left(-1\right)=6+3\)
\(\Leftrightarrow f\left(-1\right)=a.\left(-1\right)=9\)
\(\Leftrightarrow f\left(-1\right)=a=9:\left(-1\right)\)
\(\Leftrightarrow f\left(-1\right)=a=-9\)
Vậy \(f\left(-1\right)=6\) thì \(a=-9\)