Đáp án: $D$
Giải thích các bước giải:
Ta có:
$\dfrac1{x+y}+\dfrac1{y+z}+\dfrac{1}{z+x}=\dfrac3{10}$
$\to \dfrac{4000}{x+y}+\dfrac{4000}{y+z}+\dfrac{4000}{z+x}=1200$
$\to \dfrac{x+y+z}{x+y}+\dfrac{x+y+z}{y+z}+\dfrac{x+y+z}{z+x}=1200$ vì $x+y+z=4000$
$\to (1+\dfrac{z}{x+y})+(1+\dfrac{x}{y+z})+(1+\dfrac{y}{z+x})=1200$
$\to \dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}+3=1200$
$\to \dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}=1197$
$\to Q=1197$
