*Lời giải :
Ta có : `|2x - 3y| + (4y - 5z)^2 = 0`
Vì \(\left\{ \begin{array}{l}|2x-3y|≥0∀x,y\\(4y-5z)^2≥0∀x,y\end{array} \right.\)
`-> |2x - 3y| + (4y - 5z)^2 ≥ 0∀x,y,z`
Dấu "`=`" xảy ra khi :
\(\left\{ \begin{array}{l}2x-3y=0\\4y-5z=0\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}2x=3y\\4y=5z\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}\dfrac{x}{3} = \dfrac{y}{2}\\ \dfrac{y}{5} = \dfrac{z}{4}\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}\dfrac{x}{15} = \dfrac{y}{10}\\ \dfrac{y}{10} = \dfrac{z}{8}\end{array} \right.\)
`⇔ x/15 = y/10 = z/8`
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
`x/15 = y/10 = z/8 = (x + y + z)/(15 + 10 + 8) = 33/33 = 1`
`⇔` \(\left\{ \begin{array}{l}\dfrac{x}{15} = 1\\ \dfrac{y}{10}=1\\ \dfrac{z}{8}=1\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}x=15\\y=10\\z=8\end{array} \right.\)
