a) \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{5}\)
\(\Leftrightarrow\frac{2015}{a+b}+\frac{2015}{b+c}+\frac{2015}{c+a}=403\)
\(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=403\)
\(\Leftrightarrow3+\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=403\)
\(\Leftrightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=400\)
b) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Đặt \(\frac{a}{c}=\frac{b}{d}=k\Rightarrow\hept{\begin{cases}a=ck\\b=dk\end{cases}}\)
Thay vào rồi c/m nhé
a) Từ đẳng thức : \(A=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
\(\Rightarrow A+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(\Rightarrow A+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)
\(\Rightarrow A+3=\left(a+b+c\right).\frac{1}{b+c}+\left(a+b+c\right).\frac{1}{a+c}+\left(a+b+c\right).\frac{1}{a+b}\)
\(\Rightarrow A+3=\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow A+3=2015.\frac{1}{5}\)
\(\Rightarrow A+3=403\)
\(\Rightarrow A=400\)
Vậy A = 400
b) Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
Khi đó : \(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2\left(bk\right)^2-3b^2k+5b^2}{2\left(bk\right)^2+3b^2k}=\frac{2k^2b^2-3b^2k+5b^2}{2b^2k^2+3b^2k}=\frac{b^2\left(2k^2-3k+5\right)}{b^2\left(2k^2+3k\right)}\)
\(=\frac{2k^2-3k+5}{2k^2+3k}\left(1\right)\);
\(\frac{2c^2-3cd+5d^2}{2c^2+3cd}=\frac{2\left(dk\right)^2-3d^2k+5d^2}{2\left(dk\right)^2+3d^2k}=\frac{2d^2k^2-3d^2k+5d^2}{2d^2k^2+3d^2k}=\frac{d^2.\left(2k^2-3k+5\right)}{d^2\left(2k^2+3k\right)}\)
\(=\frac{2k^2-3k+5}{2k^2+3k}\left(2\right)\)
Từ (1) và (2) => \(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2c^2-3cd+5d^2}{2c^2+3cd}\)(đpcm)
