\(a.\)
Ta có : \(y=f\left(x\right)=\frac{6}{2x+1}\)
\(\Rightarrow f\left(-5\right)=\frac{6}{2.\left(-5\right)+1}=\frac{6}{-9}=-\frac{2}{3}\)
\(f\left(7\right)=\frac{6}{2.7+1}=\frac{6}{15}=\frac{2}{5}\)
\(b.\)
Ta có : \(y=f\left(x\right)=\frac{6}{2x+1}\)
\(\Rightarrow y=f\left(x\right)=10\)
\(\Rightarrow\frac{6}{2x+1}=10\)
\(\Rightarrow2x+1=6:10=0,6\)
\(\Rightarrow2x=0,6-1=-0,4\)
\(\Rightarrow x=-0,4:2=-0,2\)
Vậy : \(x=-0,2\)