Question

cho \(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}.tínhP=\dfrac{a+b}{2c}+\dfrac{b+c}{3a}+\dfrac{c+a}{4b}\)

Answer 1

Lời giải:

Nếu $a+b+c=0$ thì:

\(a+b=-c; b+c=-a; c+a=-b\)

\(\Rightarrow \left\{\begin{matrix} \frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=-1(\text{thỏa mãn giả thiết})\\ P=\frac{-c}{2c}+\frac{-a}{3a}+\frac{-b}{4b}=\frac{-1}{2}+\frac{-1}{3}+\frac{-1}{4}=\frac{-13}{12}\end{matrix}\right.\)

Nếu $a+b+c\neq 0$. Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{a+b+c}{2(a+b+c)}=\frac{1}{2}\)

\(\Rightarrow b+c=2a; c+a=2b; a+b=2c\)

\(\Rightarrow P=\frac{a+b}{2c}+\frac{b+c}{3a}+\frac{c+a}{4b}=\frac{2c}{2c}+\frac{2a}{3a}+\frac{2b}{4b}=1+\frac{2}{3}+\frac{1}{2}=\frac{13}{6}\)