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\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1.\)
\(\Rightarrow a=b=c=2014\)
Thay a,b,c vào để tính
\(\frac{b+c+d}{a}\)= \(\frac{c+d+a}{b}\)= \(\frac{d+a+b}{c}\)= \(\frac{a+b+c}{d}\)
= \(\frac{b+c+d+c+d+a+d+a+b+a+b+c}{a+b+c+d}\)
= \(\frac{3a+3b+3c+3d}{a+b+c+d}\)
= \(\frac{3\left(a+b+c+d\right)}{a+b+c+d}\)= 3
vậy k = 3
b+c+d/a=c+d+a/b=d+a+b/c=a+b+c/d=k
áp dụng tc dãy tỉ số bằng nhau ta được:
b+c+d+c+d+a+d+a+b+a+b+c/a+b+c+d=k
=>3a+3b+3c+3d/a+b+c+d=k
=>3+k
=>k=3
Vậy k=3
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\Leftrightarrow\)\(\frac{a}{b+c}+1=\frac{b}{a+c}+1=\frac{c}{a+b}+1\Leftrightarrow\)\(\frac{a+b+c}{b+c}=\frac{b+a+c}{a+c}=\frac{c+a+b}{a+b}\Leftrightarrow b+c=a+c=a+b\Leftrightarrow a=b=c\)
\(\Rightarrow P=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\frac{a+a}{a}+\frac{a+a}{a}+\frac{a+a}{a}=2+2+2=6\)
Ta có:
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\Leftrightarrow\)
\(\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}=\frac{b+c+a+c+a+b}{a+b+c}=2\)
\(\Rightarrow P=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=3.2=6\)
bài này có 2 trường hợp nhé =))
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\Rightarrow1+\frac{a}{b+c}=1+\frac{b}{a+c}=1+\frac{c}{a+b}\)
\(\Rightarrow\frac{a+b+c}{b+c}=\frac{a+b+c}{a+c}=\frac{a+b+c}{a+b}\)
\(TH1:a+b+c=0\)
\(\Rightarrow\hept{\begin{cases}b+c=-a\\a+c=-b\\a+b=-c\end{cases}\Rightarrow P=\frac{-a}{a}+\frac{-b}{b}+\frac{-c}{c}=-3}\)
\(TH2:a+b+c\ne0\)
\(\Rightarrow\hept{\begin{cases}b+c=a+c\Rightarrow a=b\\a+c=a+b\Rightarrow c=b\\a+b=b+c\Rightarrow a=c\end{cases}\Rightarrow a=b=c}\)
\(\Rightarrow P=\frac{a+a}{a}+\frac{b+b}{b}+\frac{c+c}{c}=2.3=6\)
Vậy P=-3 hay P=6
Do \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\\ \)
=> \(\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b}{c}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b}{c}=\frac{a+b+c+a+b+c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
=> \(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}=2+2+2=6\)
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)\(\Rightarrow b+c=2a\)
\(\Rightarrow a+c=2b\)
\(\Rightarrow a+b=2c\)
\(D=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\)
\(D=\frac{2a}{a}=\frac{2b}{b}=\frac{2c}{c}\)
\(D=2+2+2\)
\(D=6\)
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
=>b+c=2a
=>a+c=2b
=>a+b=2c
\(D=\frac{b+c}{a}+\frac{a+c}{b}=\frac{a+b}{c}\)
\(D=\frac{2a}{a}+\frac{2b}{b}+\frac{2c}{c}\)
\(D=2+2+2\)
D=6
Vậy D=6
^...^ 
^_^
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)(vì a+b+c khác 0)
=> a=b;b=c;c=a
=> a=b=c Mà a=3
=> a.b.c=27
Vậy a.b.c=27
a =b =c = 3
=> a.b.c = 27