ta có 3x=4y=6z [BCNN(3,4,6)= 12]
$\frac{3x}{12}$= $\frac{4y}{12}$= $\frac{6z}{12}$=>$\frac{x}{4}$= $\frac{y}{3}$= $\frac{z}{2}$=$\frac{x-y+z}{4-3+2}$ $\frac{9}{3}$ = $3^{}$
$áp^{}$ $dụng^{}$ $t/c^{}$ $dãy^{}$ $tỉ ^{}$ $số^{}$ $bằng^{}$ $nhau^{}$ $ta^{}$ $có^{}$
$do^{}$ $đó^{}$
$\frac{x}{4}$ = $3^{}$ => $x^{}$ = $3^{}$ .$4^{}$= $12^{}$ $\\$ $\frac{y}{3}$= 3=> $y^{}$= $3^{}$ .$3^{}$ =$9^{}$ $\\$ $\frac{z}{2}$= $3^{}$ =>$z^{}$= $2^{}$. $3^{}$ =$6^{}$
$vậy^{}$ $x^{}$= $12^{}$; $y^{}$= $9^{}$; $z^{}$ =$6^{}$
