Đáp án đúng:
Giải chi tiết:Ta có: \(f(x)=\,|2+x|\)
\(\Rightarrow f(x)=3\Rightarrow |2+x|=3\)
TH1: Với \(x+2\ge 0\Leftrightarrow x\ge -2\) ta có:
\(\begin{align} & x+2=3 \\ & x\,\,\,\,\,\,\,\,\,\,\,=3-2 \\ & x\,\,\,\,\,\,\,\,\,\,\,=1\,\,\,\,\left( tm \right). \\ \end{align}\)
TH2: Với \(x+2<0\Leftrightarrow x<-2\) ta có:
\(\begin{align} & -\left( x+2 \right)=3 \\ & -x-2\,\,\,\,\,\,\,=3 \\ & -x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=3+2 \\ & -x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=5 \\ & \,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-5\,\,\,\left( tm \right) \\ \end{align}\)
Vậy với \(x\in \text{ }\!\!\{\!\!\text{ -5;1 }\!\!\}\!\!\text{ }\)thì \(y=f(x)=3\).
