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\(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{b+a}=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}-3=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3=2015.\frac{1}{90}-3=19\frac{7}{18}\)
\(Q=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
=> 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
a, Đặt \(\frac{x}{5}=\frac{y}{4}=\frac{z}{3}=k\Rightarrow x=5k,y=4k,z=3k\)
Ta có: \(P=\frac{x+2y-3z}{x-2y+3z}=\frac{5k+2.4k-3.3k}{5k-2.4k+3.3k}=\frac{4k}{6k}=\frac{2}{3}\)
b, \(Q+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(Q+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)
\(Q+3=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)
\(Q+3=2015\cdot\frac{1}{5}=403\)
=>Q=403-3=400
a,\(\frac{x}{5}=\frac{y}{4}=\frac{z}{3}=k\)
\(\Rightarrow P=\frac{5k+2.4k-3.3k}{5k-2.4k+3.3k}=\frac{4}{6}=\frac{2}{3}\)
b, \(Q=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(\Rightarrow Q+3=\left(1+\frac{a}{b+c}\right)+\left(1+\frac{b}{c+a}\right)+\left(1+\frac{c}{a+b}\right)\)
\(\Rightarrow Q+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)
\(\Rightarrow Q+3=\frac{a+b+c}{b+c+c+a+a+b}=\frac{2015}{5}=403\)
\(\Rightarrow Q=400\)
Vậy Q = 400
\(S=\frac{2015-\left(a+b\right)}{a+b}+\frac{2015-\left(b+c\right)}{b+c}+\frac{2015-\left(a+c\right)}{a+c}=\frac{2015}{a+b}-\frac{a+b}{a+b}+\frac{2015}{b+c}-\frac{b+c}{b+c}+\frac{2015}{a+c}-\frac{a+c}{a+c}\)
\(S=2015.\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)-3=2015.\frac{1}{10}-3=\frac{1085}{10}\)
