a) `(3x+2)(-2/5 x -7) =0`
`=>` \(\left[ \begin{array}{l}3x+2=0\\-2/5 x -7=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}3x= -2\\-2/5 x = 7\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=-2/3\\x=7:(-2/5)\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=-2/3\\x=-35/2\end{array} \right.\)
Vậy `x= -2/3` ; `-35/2`
b) `A= 2019/1.2 + 2019/2.3 + 2019/3.4 +...+2019/2018.2019`
`A= 2019(1/1.2 + 1/2.3 + 1/3.4 +...+1/2018.2019)`
`A= 2019( 1-1/2+1/2-1/3+...+1/2018-1/2019)`
`A= 2019 ( 1-1/2019)`
`A= 2019 . 2018/2019`
`A= 2018`
Vậy `A= 2018`
