cho 3 số x,y,z thỏa mãn 101/x+y+101/y+z+101/z+x=20x/x+y+20y/y+z+20z/z+x=2020/2021.tính M=x+y+z

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1. Đáp án: $x+y+z=308.2$

   Giải thích các bước giải:

   Sửa đề:

   $\dfrac{101}{x+y}+\dfrac{101}{y+z}+\dfrac{101}{z+x}=\dfrac{20x}{y+z}+\dfrac{20y}{z+x}+\dfrac{20z}{x+y}=\dfrac{2020}{2021}$

   Ta có:

   $\dfrac{101}{x+y}+\dfrac{101}{y+z}+\dfrac{101}{z+x}=\dfrac{20x}{y+z}+\dfrac{20y}{z+x}+\dfrac{20z}{x+y}=\dfrac{2020}{2021}$

   $\to \dfrac{101}{x+y}+\dfrac{101}{y+z}+\dfrac{101}{z+x}=\dfrac{2020}{2021}$

   $\to 101(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x})=\dfrac{2020}{2021}$

   $\to \dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}=\dfrac{20}{2021}$

   Lại có:

   $\dfrac{20x}{y+z}+\dfrac{20y}{z+x}+\dfrac{20z}{x+y}=\dfrac{2020}{2021}$

   $\to \dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=\dfrac{101}{2021}$

   $\to \dfrac{x}{y+z}+1+\dfrac{y}{z+x}+1+\dfrac{z}{x+y}+1=\dfrac{101}{2021}+3$

   $\to \dfrac{x+y+z}{y+z}+\dfrac{x+y+z}{z+x}+\dfrac{z+x+y}{x+y}=\dfrac{6164}{2021}$

   $\to (x+y+z)\cdot (\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x})=\dfrac{6164}{2021}$

   $\to (x+y+z)\cdot \dfrac{20}{2021}=\dfrac{6164}{2021}$

   $\to x+y+z=308.2$

   Trả lời