Giải thích các bước giải:

Ta có:

$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{2023}{a+b}+\dfrac{2023}{b+c}+\dfrac{2023}{c+a}=\dfrac{2023}{2024}$

$\to \begin{cases} \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{2023}{2024}\\ \dfrac{2023}{a+b}+\dfrac{2023}{b+c}+\dfrac{2023}{c+a}=\dfrac{2023}{2024}\end{cases}$

$\to \begin{cases} \dfrac{a}{b+c}+1+\dfrac{b}{c+a}+1+\dfrac{c}{a+b}+1=3+\dfrac{2023}{2024}\\ \dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{2024}\end{cases}$

$\to \begin{cases} \dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}=\dfrac{8095}{2024}\\ \dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{2024}\end{cases}$

$\to \begin{cases} (a+b+c)\cdot (\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a})=\dfrac{8095}{2024}\\ \dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{2024}\end{cases}$

$\to (a+b+c)\cdot dfrac1{2024}=\dfrac{8095}{2024}$

$\to a+b+c=8095$