\(\frac{2015}{a+b}+\frac{2015}{b+c}+\frac{2015}{c+a}=\frac{2015}{90}\)

\(\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{2015}{90}\)

\(1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{2015}{90}\)

\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{b+a}=\frac{2015}{90}-3=\frac{349}{18}\)

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

=> Q + 3 = \(\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)\)

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

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

\(=2015.\frac{1}{5}=403\)\(\text{Vì }\hept{\begin{cases}a+b+c=2015\\\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{5}\end{cases}}\)

Khi đó Q = 3 = 403

=> Q = 400

Vậy Q = 400

lớp 6 mà

lớp 9 đó

\(a)\) Ta có : 

\(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=\frac{1}{2015}\)

\(\Leftrightarrow\)\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=\left(a+b+c\right).\frac{1}{2015}\)

\(\Leftrightarrow\)\(\frac{a+b+c}{a+b}+\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=\frac{a+b+c}{2015}\)

\(\Leftrightarrow\)\(1+\frac{c}{a+b}+1+\frac{b}{a+c}+1+\frac{a}{b+c}=\frac{2015}{2015}\)

\(\Leftrightarrow\)\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1-3\)

\(\Leftrightarrow\)\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=-2\)

Vậy ...

a²⁰¹⁴+b²⁰¹⁴+c²⁰¹⁴=1

a²⁰¹⁵⁺b²⁰¹⁵+c²⁰¹⁵=1

=>a²⁰¹⁴+b²⁰¹⁴+c²⁰¹⁴=a²⁰¹⁵⁺b²⁰¹⁵+c²⁰¹⁵=1

=>a=b=1

=>A=3

đây là hướng giải thôi nhé

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