Đáp án:
$\text{ x = 1515 , y = 2020 , z = 2525 }$
Giải thích các bước giải:
$\dfrac{4x-3y}{5}=$ $\dfrac{5y-4z}{4}=$ $\dfrac{3z-5x}{4}$
$=\dfrac{5(4x-3y+3(5y-4z)+4(3z-5x)}{5.5+4.4+3.3}$
$=\dfrac{20x-15y+15y-12z+12z-20x}{50}=$ $\dfrac{0}{50}=0$
$⇔\left[\begin{array}{ccc}\dfrac{4x-3y}{5}=0\\\dfrac{5y-4z}{5}=0\\\dfrac{3z-5x}{5}=0\end{array}\right]$ $⇒\left[\begin{array}{ccc}4x-3y=0\\5y-4z=0\\3z-5x=0\end{array}\right]$ $⇒\left[\begin{array}{ccc}\dfrac{x}{3}= \dfrac{y}{4}\\\dfrac{y}{4}= \dfrac{z}{5}\\\dfrac{z}{5}= \dfrac{x}{3}\end{array}\right]$
theo tính chất của dãy tỉ số bằng nhau ta có :
$\dfrac{x}{3}=$ $\dfrac{y}{4}=$ $\dfrac{z}{5}=$ $\dfrac{x-y+z}{3-4+5}=$ $\dfrac{2020}{4}=505$
$⇒\dfrac{x}{3}=505⇒x=3.505=1515$
$⇒\dfrac{y}{4}=505⇒y=4.505=2020$
$⇒\dfrac{z}{5}=505⇒z=5.505=2525$
