*Lời giải :
`1)`
`2x = y/3 = z/5`
`⇔` \(\left\{ \begin{array}{l}2x=\dfrac{y}{3}\\2x=\dfrac{z}{5}\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}2x.3=y\\2x.5=z\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}6x=y (1)\\10x=z (2)\end{array} \right.\)
Ta có : `x + y -z/2 = -20`
`⇔ x + 6x - (10x)/2 = -20`
`⇔ x + 6x - 5x = -20`
`⇔ 2x = -20`
`⇔ x = -10`
Với `x = -10`
Từ `(1) → 6 . (-10) = y -> y = -60`
Từ `(2) → 10 . (-10) = z -> z= -100`
Vậy `x = -10, y = -60,z=-100`
`2)`
Đặt `f (x) = (3 - 4x + x^2)^{2016} . (3 + 4x + x^2)^{2017}`
Để tính tổng `-> x = 1`
`-> f (x) = (3 - 4 . 1 + 1^2)^{2016} . (3 + 4 . 1 + 1^2)^{2017}`
`-> f (x) = (3- 4 + 1)^{2016} . (3 + 4 + 1)^{2017}`
`-> f (x) = 0^{2016} . 8^{2017}`
`-> f (x) = 0`
