Áp dụng t/c dãy tỉ số bằng nhau :
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b+b+c+c+a}{c+a+b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)(a,b,c # 0 nên a + b + c # 0 )
Từ \(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)
=> \(\dfrac{c}{a+b}=\dfrac{a}{b+c}=\dfrac{b}{c+a}\)
Áp dụng ....
\(\dfrac{c}{a+b}=\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c+a+b}{a+b+b+c+c+a}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)(a + b + c # 0 )
Ta có : \(A=\dfrac{a}{b+c}+\dfrac{a+b}{c}\)
\(A=\dfrac{1}{2}+2\)
\(A=\dfrac{5}{2}\)
Vậy \(A=\dfrac{5}{2}\)
Đề đâu có nói a;b;c âm hay dương,nên \(a+b+c=0\) vẫn được nhé
Lời giải:
Với \(a+b+c=0\) ta có: \(\left\{{}\begin{matrix}a=-\left(b+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\Leftrightarrow A=\dfrac{a}{-a}+\dfrac{-c}{c}=\left(-1\right)+\left(-1\right)=-2\)
Với \(a+b+c\ne0\) áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b+b+c+c+a}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\Leftrightarrow A=\dfrac{a}{2a}+\dfrac{2c}{c}=\dfrac{1}{2}+2=\dfrac{5}{2}\)
ta co:\(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}=\dfrac{a+b+b+c+c+a}{c+a+b}=\dfrac{2.\left(a+b+c\right)}{a+b+c}=2\)=>\(\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)
Thay vao A=\(\dfrac{a}{2a}+\dfrac{2c}{c}=\dfrac{1}{2}+2=\dfrac{5}{2}\)
