Đặt `a/2 = (3b)/5 = (2c)/7 = d/4 =k`
`=> a= 2k; b= (5k)/3 ; c= (7k)/2 ; d= 4k`
Ta có: `a+b+c+ d=36`
`=> 2k + 5/3 k + 7/2 k + 4k =36`
`=> k(2 + 5/3 + 7/2 +4) = 36`
`=>67/6 k = 36`
`=>k = 216/67`
+) `a= 2. 216/67= 432/67`
+) `b= 5/3 . 216/67= 360/67`
+) `c= 7/2 . 216/67= 756/67`
+) `d= 4. 216/67= 864/67`
Vậy `a= 432/67 ; b= 360/67 ; c= 756/67 ;d = 864/67`