\(\left(3x+1\right)^2:\left(-\dfrac{1}{4}\right)=\left(\dfrac{-49}{4}\right)\)
\(\Rightarrow\left(3x+1\right)^2=\dfrac{49}{16}\)
\(\Rightarrow\left[{}\begin{matrix}3x+1=\dfrac{7}{4}\\3x+1=\dfrac{-7}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{-11}{12}\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{4}\) hoặc \(x=\dfrac{-11}{12}\)
\(\left(3x+1\right)^2:\left(-\dfrac{1}{4}\right)=\dfrac{-49}{4}\)
\(\Rightarrow\left(3x+1\right)^2=\dfrac{-49}{-16}\)
\(\Rightarrow\left(3x+1\right)^2=\dfrac{49}{16}\)
\(\Rightarrow\left[{}\begin{matrix}3x+1=\dfrac{7}{3}\\3x+1=\dfrac{-7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{9}\\x=\dfrac{-10}{9}\end{matrix}\right.\)
Vậy \(x=\dfrac{4}{9}\) hoặc \(x=\dfrac{-10}{9}\)
\(\left(3x+1\right)^2:\left(\dfrac{-1}{4}\right)=\dfrac{-49}{4}\)
\(\Rightarrow\left(3x+1\right)^2=\dfrac{49}{16}\)
\(\Rightarrow\left[{}\begin{matrix}3x+1=\dfrac{7}{4}\\3x+1=\dfrac{-7}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{-11}{12}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-11}{12};\dfrac{1}{4}\right\}\).
