Đáp án: $M=308.2$
Giải thích các bước giải:
Ta có:
$\dfrac{101}{x+y}+\dfrac{101}{y+z}+\dfrac{101}{z+x}=\dfrac{2020}{2021}$
$\to (\dfrac1{x+y}+\dfrac1{y+z}+\dfrac1{z+x})\cdot 101=\dfrac{2020}{2021}$
$\to \dfrac1{x+y}+\dfrac1{y+z}+\dfrac1{z+x}=\dfrac{20}{2021}$
Ta có:
$\dfrac{20x}{y+z}+\dfrac{20y}{z+x}+\dfrac{20z}{x+y}=\dfrac{2020}{2021}$
$\to \dfrac{20(x+y+z)-20(y+z)}{y+z}+\dfrac{20(x+y+z)-20(z+x)}{z+x}+\dfrac{20(x+y+z)-20(x+y)}{x+y}=\dfrac{2020}{2021}$
$\to \dfrac{20(x+y+z)}{y+z}-20+\dfrac{20(x+y+z)}{z+x}-20+\dfrac{20(x+y+z)}{x+y}-20=\dfrac{2020}{2021}$
$\to \dfrac{20(x+y+z)}{y+z}+\dfrac{20(x+y+z)}{z+x}+\dfrac{20(x+y+z)}{x+y}-60=\dfrac{2020}{2021}$
$\to 20(x+y+z)\cdot (\dfrac1{x+y}+\dfrac1{y+z}+\dfrac1{z+x})-60=\dfrac{2020}{2021}$
$\to 20(x+y+z)\cdot \dfrac{20}{2021}-60=\dfrac{2020}{2021}$
$\to (x+y+z)\cdot \dfrac{400}{2021}=\dfrac{2020}{2021}+60$
$\to x+y+z=308.2$
