a)
Ta có :
`5x = 3y`
⇒ ` \frac{x}{3} = \frac{y}{5} = \frac{x-y}{3-5} = \frac{-5}{-2} = \frac{5}{2}`
⇒ $x = \dfrac{15}{2} ; y = \dfrac{25}{2}$
Vậy `(x ; y ) = (\frac{15}{2} ; \frac{25}{2})`
b)
Ta có :
$2x = 3y ⇔ \dfrac{x}{3} = \dfrac{y}{2} ⇔ \dfrac{x}{21} = \dfrac{y}{14}$
$5y = 7z ⇔ \dfrac{y}{7} = \dfrac{z}{5} ⇔ \dfrac{y}{14} = \dfrac{z}{10}$
⇒ $\dfrac{x}{21} = \dfrac{y}{14} = \dfrac{z}{10}$
⇒ $\dfrac{3x}{63} = \dfrac{5y}{70} = \dfrac{7z}{70} = \dfrac{3x - 5y + 7z}{63 - 70 + 70} = \dfrac{30}{63} = \dfrac{10}{21}$
⇒ $x = \dfrac{210}{21} = 10$
$y = \dfrac{140}{21}$
$z = \dfrac{100}{21}$
Vậy `(x ; y ; z) = (10 ; \frac{140}{21} ; \frac{100}{21})`
#Winner112
