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:
$\dfrac{20x}{y+z}+\dfrac{20y}{z+x}+\dfrac{20z}{x+y}=\dfrac{2020}{2021}$
$\to \dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=\dfrac{101}{2021}$
$\to \dfrac{x}{y+z}+1+\dfrac{y}{z+x}+1+\dfrac{z}{x+y}+1=\dfrac{101}{2021}+3$
$\to \dfrac{x+y+z}{y+z}+\dfrac{x+y+z}{z+x}+\dfrac{z+x+y}{x+y}=\dfrac{6164}{2021}$
$\to (x+y+z)\cdot (\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x})=\dfrac{6164}{2021}$
$\to (x+y+z)\cdot \dfrac{20}{2021}=\dfrac{6164}{2021}$
$\to x+y+z=308.2$
