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\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
\(\Rightarrow\frac{a+b-c}{c}+1=\frac{b+c-a}{a}+1=\frac{c+a-b}{b}+1\)
\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\)
+)Nếu a+b+c=0\(\Rightarrow a+b=-c;b+c=-a;c+a=-b\)
\(\Rightarrow B=\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}=\frac{-c}{a}.\frac{-b}{c}.\frac{-a}{b}=\frac{-\left(abc\right)}{abc}=-1\)
Nếu \(a+b+ c\ne0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow a+b=2c\)
\(b+ c=2a\)
\(c+a=2b\)
\(\Rightarrow B=\frac{2c}{a}.\frac{2b}{c}.\frac{2a}{b}=2.2.2=8\)
a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)
Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)
b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)
\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)
\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)
\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)
\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)
Do đó: +) \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)
+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)
+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)
#)Giải :
\(\frac{a+b-c}{c}=\frac{a+c-b}{b}=\frac{b+c-a}{a}\)
\(\Leftrightarrow\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}\)
TH1 : \(a+b+c=0\Leftrightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}\Leftrightarrow M=\frac{\left(-c\right)\left(-a\right)\left(-b\right)}{abc}=-1}\)
TH2 : \(a+b+c\ne0\)
Áp dụng tính chất dãy tỉ số bằng nhau :
\(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}=\frac{a+b-c+a-b+c-a+b+c}{c+b+a}=1\)
\(\Rightarrow\hept{\begin{cases}a+b-c=c\\a-b+c=b\\-a+b+c=a\end{cases}\Rightarrow\hept{\begin{cases}a+b=2c\\a+c=2b\\b+c=2a\end{cases}\Rightarrow}M=\frac{2c.2b.2a}{abc}=8}\)
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\)
\(\Rightarrow\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}=\dfrac{1+1+1}{a+b+c}=\dfrac{3}{a+b+c}=\dfrac{3}{1}=3\)
\(\Rightarrow a=b=c=\dfrac{1}{3}\)
\(\Rightarrow A=\dfrac{a^3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=a^3=\left(\dfrac{1}{3}\right)^3=\dfrac{1}{27}\)