*Lời giải :
Ta có : `|4x - 3y| + |6y - 5z| = 0`
Vì \(\left\{ \begin{array}{l}|4x-3y|≥0∀x,y\\|6y-5z|≥0∀y,z\end{array} \right.\)
`-> |4x - 3y| + |6y - 5z| ≥0∀x,y,z`
Dấu " `=`" xảy ra khi :
\(\left\{ \begin{array}{l}4x - 3y = 0\\6y - 5z = 0\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}4x = 3y\\6y = 5z\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}\dfrac{x}{3} = \dfrac{y}{4}\\ \dfrac{y}{5} = \dfrac{z}{6}\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}\dfrac{x}{15} = \dfrac{y}{20}\\ \dfrac{y}{20} = \dfrac{z}{24}\end{array} \right.\)
`⇔ x/15 = y/20 = z/24`
Đặt `x/15 = y/20 = z/24 = k`
`⇔` \(\left\{ \begin{array}{l}x=15k\\y=20k\\z=24k\end{array} \right.\) (*)
Thay (*) vào `A = (2x + 3y + 6z)/(3x + 4y + 5z)` ta được :
`A = (2 . 15k + 3 . 20k + 6 . 24k)/(3 . 15k + 4 . 20k + 5 . 24k)`
`-> A = (30k + 60k + 144k)/(45k + 80k + 120k)`
`-> A = (k (30 + 60 + 144) )/(k (45 + 80 + 120) )`
`-> A = 234/245`
