Question

b) ( 3x+1)(7+2x)=0

Answer 1

`(3x+1)(7+2x)=0` 

`<=>` $\left[\begin{matrix} 3x+1=0\\ 7+2x=0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=\dfrac{-1}{3}\\ x=\dfrac{-7}{2}\end{matrix}\right.$

Vậy `S={[-1]/3;[-7]/2}`

Answer 2

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

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

Answer 3

`<=>[(3x+1=0),(7+2x=0):}`

`<=>[(3x=-1),(2x=-7):}`

`<=>[(x=-1/3),(x=-7/2):}`

Answer 4

\(\left(3x+1\right)\left(7+2x\right)=0\)

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