Question

Tìm nghiệm (3x-1)(4x+3)/2=(1-3x)(2x-5)/6

Answer 1

\(\dfrac{\left(3x-1\right)\left(4x+3\right)}{2}=\dfrac{\left(1-3x\right)\left(2x-5\right)}{6}\)

\(\Leftrightarrow3\left(3x-1\right)\left(4x+3\right)-\left(1-3x\right)\left(2x-5\right)=0\)

\(\Leftrightarrow3\left(3x-1\right)\left(4x+3\right)+\left(3x-1\right)\left(2x-5\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(12x+9+2x-5\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(14x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\14x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{-2}{7}\end{matrix}\right.\)

Vậy, \(S=\left\{\dfrac{1}{3};\dfrac{-2}{7}\right\}\)