Ta có: \(y=f\left(x\right)=\left|3x-2\right|\)
+) Thay \(f\left(x\right)=0\)
\(\Rightarrow0=\left|3x-2\right|\)
\(\Rightarrow3x-2=0\)
\(\Rightarrow x=\frac{2}{3}\)
Vậy \(x=\frac{2}{3}\)
+) Thay \(f\left(x\right)=\frac{1}{4}\)
\(\Rightarrow\left|3x-2\right|=\frac{1}{4}\)
\(\Rightarrow3x-2=\pm\frac{1}{4}\)
+ \(3x-2=\frac{1}{4}\)
\(\Rightarrow3x=\frac{9}{4}\)
\(\Rightarrow x=\frac{3}{4}\)
+ \(3x-2=\frac{-1}{4}\)
\(\Rightarrow3x=\frac{7}{4}\)
\(\Rightarrow x=\frac{7}{12}\)
Vậy \(x\in\left\{\frac{3}{4};\frac{7}{12}\right\}\)
+) Thay \(f\left(x\right)=2012\)
\(\Rightarrow\left|3x-2\right|=2012\)
\(\Rightarrow3x-2=\pm2012\)
+ \(3x-2=2012\Rightarrow3x=2014\Rightarrow x=\frac{2014}{3}\)
+ \(3x-2=-2012\Rightarrow3x=-2010\Rightarrow x=\frac{-2010}{3}\)
Vậy \(x\in\left\{\frac{2014}{3};\frac{-2010}{3}\right\}\)
Ta có: y = f(x) = |3x - 2|
- Với f(x) = 0 thì:
|3x - 2| = 0
=> 3x - 2 = 0
=> 3x = 0 + 2
=> 3x = 2
=> x = \(\frac{2}{3}\)
- Với f(x) = \(\frac{1}{4}\) thì:
|3x - 2| = \(\frac{1}{4}\)
\(\Rightarrow\left\{\begin{matrix}3x-2=\frac{1}{4}\\3x-2=-\frac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}3x=\frac{1}{4}+2=\frac{9}{4}\\3x=-\frac{1}{4}+2=\frac{7}{4}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=\frac{9}{4}\div3=\frac{3}{4}\\x=\frac{7}{4}\div3=\frac{7}{12}\end{matrix}\right.\)
- Với f(x) = 2012 thì:
|3x - 2| = 2012
\(\Rightarrow\left\{\begin{matrix}3x-2=2012\\3x-2=-2012\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}3x=2012+2=2014\\3x=-2012+2=-2010\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=\frac{2014}{3}\\x=-\frac{2010}{3}=-670\end{matrix}\right.\)
