Cho 4x-3y / 5 = 5y-4z/3 = 3z-5x/4 và x-y+z= 2020. Tìm x,y,z.

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 :

$\dfrac{x}{3}=$ $\dfrac{y}{4}=$ $\dfrac{z}{5}=$ $\dfrac{x-y+z}{3-4+5}=$ $\dfrac{2020}{4}=505$

$⇒\dfrac{x}{3}=505⇒x=3.505=1515$

$⇒\dfrac{y}{4}=505⇒y=4.505=2020$

$⇒\dfrac{z}{5}=505⇒z=5.505=2525$