Ta có :
`x/2=(2y)/5=(3z)/7`
`=> (2x)/4=(5y)/(12,5)=(3z)/7=(2x+5y+3z)/(4+12,5+7)=(2x+5y+3z)/(23,5)` `(1)`
Lại có :
`x/2=(2y)/5=(2z)/7`
`=> (7x)/14=y/(2,5)=`$\dfrac{5z}{\dfrac{35}{3}}=\dfrac{7x+y-5z}{14+2,5-\dfrac{35}{3}}=\dfrac{7x+y-5z}{\dfrac{29}{6}}$ `(2)`
Từ `(1), (2)`
`⇒ (2x+5y+3z)/(23,5)=`$\dfrac{7x+y-5z}{\dfrac{29}{6}}$
$⇔ A=\dfrac{2x+5y-3z}{7x+y-5z}=\dfrac{9,5}{\dfrac{29}{6}}=\dfrac{57}{29}$
Vậy `A=57/29`
Xin hay nhất !
