Áp dụng t/c dtsbn:

\(\dfrac{1}{a+b}=\dfrac{2}{b+c}=\dfrac{3}{c+a}=\dfrac{1+2+3}{2\left(a+b+c\right)}=\dfrac{6}{2\left(a+b+c\right)}=\dfrac{3}{a+b+c}\)

\(\Rightarrow\left\{{}\begin{matrix}3a+3b=a+b+c\\3b+3c=2a+2b+2c\\3a+3c=3a+3b+3c\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}c=2a\\b=0\end{matrix}\right.\)

\(Q=\dfrac{a+2021b+c}{a+2022b+c}=\dfrac{a+2a}{a+2a}=1\)

\(\frac{1}{a+b}=\frac{2}{b+c}=\frac{3}{c+a}=\frac{1+2+3}{2\left(a+b+c\right)}=\frac{3}{a+b+c}.\)

\(\Rightarrow\frac{3}{c+a}=\frac{3}{a+b+c}\Rightarrow c+a=a+b+c\Rightarrow b=0\)

\(\Rightarrow Q=\frac{a+2021b+c}{a+2022b+c}=\frac{a+c}{a+c}=1\)

rồi sao

\(\frac{1}{a+b}=\frac{1}{b+c}=\frac{1}{c+c}\Rightarrow\frac{1}{a+b}=\frac{1}{b+c}\Rightarrow a+b=b+c\)

\(\Rightarrow a=c\left(1\right)\)

\(\frac{1}{b+c}=\frac{1}{c+c}\Rightarrow b+c=c+c\Rightarrow c=b\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow a=b=c\)

\(Q=\frac{a+2021b+c}{a+2022b+c}=\frac{a+2021a+a}{a+2022a+a}\)

\(Q=\frac{a.\left(1+2021+1\right)}{a.\left(1+2022+1\right)}=\frac{2023}{2024}\)

Vậy, \(Q=\frac{2023}{2024}\)

Từ \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\) suy ra \(\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)

\(\Rightarrow\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}\\\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\\\frac{1}{c}+\frac{1}{a}=\frac{1}{a}+\frac{1}{b}\end{cases}}\)\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)

Khi đó \(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)

Câu hỏi của Đậu Đình Kiên - Toán lớp 7 - Học toán với OnlineMath