`text{Câu 1}`
`1/(x(x + 1)(x + 2)) = A/x + B/(x + 1) + C/(x + 2)`
`text{Áp dụng đồng nhất hệ số, ta có}`
`A(x + 1)(x + 2) + Bx(x + 2) + Cx(x + 1)`
`= A(x^2 + 3x + 2) + Bx^2 + 2Bx + Cx^2 + Cx`
`= Ax^2 + 3Ax + 2A + Bx^2 + 2Bx + Cx^2 + Cx`
`= (A + B + C)x^2 + (3A + 2B + C)x + 2A`
`text{Theo giả thiết}`
`->` \(\left\{ \begin{array}{l}A + B + C = 0\\3A + 2B + C = 0\\2A = 1\end{array} \right.\)
`->` \(\left\{ \begin{array}{l}A = \dfrac{1}{2}\\B = -1\\C = \dfrac{1}{2}\end{array} \right.\)
`text{Bài 2}`
`S_{n} = 1/(1.2.3) + 1/(2.3.4) + ... + 1/(n(n + 1)(n + 2))`
`-> 2S_{n} = 2/(1.2.3) + 2/(2.3.4) + ... + 2/(n(n + 1)(n + 2))`
`-> 2S_{n} = 1/(1.2) - 1/(2.3) + 1/(2.3) - 1/(3.4) + 1/(3.4) + ... + 1/(n(n + 1)) - 1/((n + 1)(n + 2))`
`-> 2S_{n} = 1/4 - 1/((n + 1)(n + 2))`
`-> 2S_{2020} = 1/4 - 1/(2021.2022) ≈ 0,25`
