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(3x + 2) × (-2/5 x -70 = 0 A = 2019/1 × 2 + 2019/2 ×3 + 2019/3 × 4 +…+2019/2018 × 2019
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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`
A=$\frac{2019}{1×2}$+ $\frac{2019}{2×3}$+$\frac{2019}{3×4}$+…+$\frac{2019}{2018×2019}$
=2019x($\frac{1}{1×2}$+$\frac{1}{2×3}$+$\frac{1}{3×4}$+…+$\frac{1}{2018×2019}$
=2019x(1-$\frac{1}{2}$+$\frac{1}{2}$-$\frac{1}{3}$+$\frac{1}{3}$-$\frac{1}{4}$+…+$\frac{1}{2018}$-$\frac{1}{2019}$)
=2019x(1-$\frac{1}{2019}$)
=2019x$\frac{2018}{2019}$
=2018
(3x+2) x (-2/5x-10)=0
⇒3x+2 = 0 hoặc -2/5x-10=0
⇒x=2/3 hoặc x=-25