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\(\frac{ab}{a+b}=\frac{bc}{b+c}\Rightarrow\frac{a}{a+b}=\frac{c}{b+c}\Rightarrow a\left(b+c\right)=c\left(a+b\right)\)
\(\Rightarrow ab+ac=ac+bc\Rightarrow a=c\)
\(\frac{bc}{b+c}=\frac{ac}{a+c}\Rightarrow\frac{b}{b+c}=\frac{a}{a+c}\Rightarrow b\left(a+c\right)=a\left(b+c\right)\)
\(\Leftrightarrow ab+bc=ab+ac\Rightarrow a=b\)
=> a=b=c
\(A=\frac{a^3+b^3+c^3}{a^2b+b^2c+c^2a}=\frac{3a^3}{3a^3}=1\)
Áp dụng tính chất của dãy tỉ số bằng nhau,ta có :
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ac}{a+c}=\frac{ab-bc}{\left(a+b\right)-\left(b+c\right)}=\frac{bc-ac}{\left(b+c\right)-\left(a+c\right)}=\frac{ab-ac}{\left(a+b\right)-\left(a+c\right)}\)
\(\Rightarrow\)a = b = c
\(\Rightarrow A=\frac{a^3+b^3+c^3}{a^2b+b^2c+c^2a}=1\)
Có: \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ac}{a+c}\)
\(\Rightarrow\frac{abc}{ac+bc}=\frac{abc}{ab+ac}=\frac{abc}{ab+bc}\)
\(\Rightarrow ac+bc=ab+ac=ab+bc\)
\(\Rightarrow\hept{\begin{cases}ac+bc=ab+ac\\ab+ac=ab+bc\\ac+bc=ab+bc\end{cases}\Rightarrow\hept{\begin{cases}bc=ab\\ac=bc\\ac=ab\end{cases}\Rightarrow}\hept{\begin{cases}a=c\\a=b\\b=c\end{cases}}}\)
\(\Rightarrow a=b=c\)(1)
Thay (1) vào A, ta được: \(A=\frac{a^3+a^3+a^3}{a^2.a+a^2.a+a^2.a}=\frac{a^3+a^3+a^3}{a^3+a^3+a^3}=1\)
Vậy A = 1
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: +) \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)
+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)
+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)
