cho 4x-3y/5 = 5y-4z/3 = 3z-5x/4 và x-y+z=2020 tìm x, y, z

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1. Đáp án:

   $\left\{\begin{matrix}x=505.3=1515& \\y=505.4=2020&\\z=505.5=2525& \end{matrix}\right.$ 

   Giải thích các bước giải:

   `(4x-3y)/5=(5y-4z)/3=(3z-5x)/4`

   `=> (20x-15y)/25=(15y-12z)/9=(12z-20x)/16` 

   `=(20x-15y+15y-12z+12z-20x)/(25+9+16)=0/50=0`

   `=>` $\left\{\begin{matrix}\dfrac{4x-3y}{5}=0& \\\dfrac{5y-4z}{3}=0&\\\dfrac{3z-5x}{4}=0& \end{matrix}\right.$ `=>` $\left\{\begin{matrix}4x-3y=0& \\5y-4z=0&\\3z-5x=0& \end{matrix}\right.$ `=>` $\left\{\begin{matrix}4x=3y& \\5y=4z&\\3z=5x& \end{matrix}\right.$

   `=>` $\left\{\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}& \\\dfrac{y}{4}=\dfrac{z}{5}&\\\dfrac{z}{5}=\dfrac{x}{3}& \end{matrix}\right.$  `=> x/3=y/4=z/5`

   `=(x-y+z)/(3-4+5)=2020/4=505`

   `=>` $\left\{\begin{matrix}x=505.3=1515& \\y=505.4=2020&\\z=505.5=2525& \end{matrix}\right.$ 

    

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2. Đáp án:

   $x=1515,y=2020,z=2525$

   Giải thích các bước giải:

   Theo tính chất dãy tỉ số bằng nhau ta có:

   $\dfrac{4x-3y}{5}=\dfrac{5y-4z}{3}=\dfrac{3z-5x}{4}=\dfrac{5(4x-3y)+3(5y-4z)+4(3z-5x)}{5.5+3.3+4.4}$

   $=\dfrac{20x-15y+15y-12z+12z-20x}{50}=\dfrac{0}{50}=0$

   $\Rightarrow \left\{\begin{array}{l} \dfrac{4x-3y}{5}=0\\ \dfrac{5y-4z}{3}=0\\ \dfrac{3z-5x}{4}=0\end{array} \right.\Rightarrow \left\{\begin{array}{l} 4x-3y=0\\ 5y-4z=0\\ 3z-5x=0\end{array} \right.\Rightarrow \left\{\begin{array}{l} \dfrac{x}{3}=\dfrac{y}{4}\\ \dfrac{y}{4}=\dfrac{z}{5}\\ \dfrac{z}{5}=\dfrac{x}{3}\end{array} \right.$

   $\Rightarrow \dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{x-y+z}{3-4+5}=\dfrac{2020}{4}=505$ (tính chất dãy tỉ số bằng nhau)

   $\Rightarrow \dfrac{x}{3}=505\Rightarrow x=1515$

   $\dfrac{y}{4}=505\Rightarrow 2020$

   $\dfrac{z}{5}=505\Rightarrow z=2525$

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