Bài làm:
Áp dụng t/c dãy tỉ số bằng nhau:
\(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}=\frac{5\left(a+b+c\right)}{a+b+c}=5\)
\(\Rightarrow\hept{\begin{cases}3a+b+c=5a\\a+3b+c=5b\\a+b+3c=5c\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b+c=3a\\a+b+c=3b\\a+b+c=3c\end{cases}}\Rightarrow a=b=c\)
Vậy \(P=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=\frac{2c}{c}+\frac{2a}{a}+\frac{2b}{b}=2+2+2=6\)
Vậy P = 6
Vì a ; b ; c > 0 => a + b + c > 0
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}=\frac{3a+b+c+a+3b+c+a+b+3c}{a+b+c}\)
\(=\frac{5\left(a+b+c\right)}{a+b+c}=5\)
\(\Rightarrow\hept{\begin{cases}3a+b+c=5a\\a+3b+c=5b\\a+b+3c=5c\end{cases}}\Rightarrow\hept{\begin{cases}b+c=2a\\a+c=2b\\a+b=2c\end{cases}}\)
Khi đó P = \(\frac{2c}{c}+\frac{2a}{a}+\frac{2b}{b}=2+2+2=6\)
\(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}\)
\(=\frac{3a}{a}+\frac{b+c}{a}=\frac{3b}{b}+\frac{a+c}{b}=\frac{3c}{c}+\frac{a+b}{c}\)
\(=3+\frac{b+c}{a}=3+\frac{a+c}{b}=3+\frac{a+b}{c}\)
\(=\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}\)
theo tính chất dãy tỉ số bằng nhau
\(\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}=\frac{b+c+a+c+a+b}{a+b+c}=\frac{2a+2b+2c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\hept{\begin{cases}\frac{b+c}{a}=2\Leftrightarrow b+c=2a\left(1\right)\\\frac{a+c}{b}=2\Leftrightarrow a+c=2b\left(2\right)\\\frac{a+b}{c}=2\Leftrightarrow a+b=2c\left(3\right)\end{cases}}\)
thay (1);(2);(3) vào \(P=\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}\)ta được
\(P=\frac{2c}{c}+\frac{2a}{a}+\frac{2b}{b}=2+2+2=6\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}=\frac{3a+b+c+a+3b+c+a+b+3c}{a+b+c}=\frac{5\left(a+b+c\right)}{a+b+c}=5\)
\(\Rightarrow\hept{\begin{cases}3a+b+c=5a\\a+3b+c=5b\\a+b+3c=5c\end{cases}}\Rightarrow\hept{\begin{cases}b+c=2a\\a+c=2b\\a+b=2c\end{cases}}\Rightarrow a=b=c\)
Thế vào P ta được :
\(P=\frac{c+c}{c}+\frac{a+a}{a}+\frac{b+b}{b}=\frac{2c}{c}+\frac{2a}{a}+\frac{2b}{b}=2+2+2=6\)
Sử dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}=\frac{3a+b+c+a+3b+c+a+b+3c}{a+b+c}\)
\(=\frac{3\left(a+b+c\right)+2\left(a+b+c\right)}{a+b+c}=\frac{5\left(a+b+c\right)}{a+b+c}=5\)
\(< =>\hept{\begin{cases}\frac{3a+b+c}{a}=5\\\frac{3b+a+c}{b}=5\\\frac{3c+a+b}{c}=5\end{cases}< =>\hept{\begin{cases}3a+b+c=5a\\3b+a+c=5b\\3c+a+b=5c\end{cases}}}\)
\(< =>\hept{\begin{cases}a+b+c=5a-2a=3a\\a+b+c=5b-2b=3b\\a+b+c=5c-2c=3c\end{cases}}\)
\(< =>a+b+c=3a=3b=3c< =>a=b=c\)
Khi đó \(P=\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}=\frac{a+a}{a}+\frac{a+a}{a}+\frac{a+a}{a}\)
\(=\frac{2a}{a}+\frac{2a}{a}+\frac{2a}{a}=\frac{2a+2a+2a}{a}=\frac{6a}{a}=6\)
Vậy \(P=6\)
Áp dụng dãy tỉ số bằng nhau :
\(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}=\frac{5\left(a+b+c\right)}{a=b+c}=5\)
\(\Rightarrow\hept{\begin{cases}3a+b+c=5a\\a+3b+c=5b\\a+b+3c=5c\end{cases}\Leftrightarrow\hept{\begin{cases}a+b+c=3a\\a+b+c=3b\Rightarrow\\a+b+c=3c\end{cases}}a=b=c}\)
Vậy \(P=\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=\frac{2a}{c}+\frac{2a}{a}+\frac{2b}{b}=2+2+2=6\)
Vậy \(P=6\)
