Ta có :
`2x = -3y = 4z`
`=>` $\dfrac{2x}{12}$ = $\dfrac{-3y}{12}$ = $\dfrac{4z}{12}$
`=> x/6 = y/-4 = z/3`
Đặt `: x/6 = y/-4 = z/3 = n`
Ta có :
`x/6 = n => x = 6n`
`y/-4 = n => y = -4n`
`z/3 = n => z = 3n`
Theo đề bài , ta có :
`1/x + 1/y + 1/z = 3`
`=>` $\dfrac{1}{6n}$ + $\dfrac{1}{-4n}$ + $\dfrac{1}{3n}$ `= 3`
`=> 1/6 . 1/n + 1/-4 . 1/n + 1/3 . 1/n = 3`
`=> ( 1/6 + 1/-4 + 1/3 ) . 1/n = 3`
`=> 1/4 . 1/n = 3`
`=>` $\dfrac{1}{4n}$ `= 3`
`=> n = 1/12`
Ta có `: n = 1/12`
`=> x = 6 . 1/12 = 6/12 = 1/2`
`y = -4 . 1/12 = -1/3`
`z = 3 . 1/12 = 1/4`
Vậy `: x = 1/2 ; y = -1/3 ; z = 1/4`
