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\(a-b=13\Rightarrow a=b+13\)
thay \(a=b+13\) vào biểu thức thì ta có:
\(\frac{3a-b}{2a+13}-\frac{3b-a}{2b-13}=\frac{3\left(b+13\right)-b}{2\left(b+13\right)+13}-\frac{3b-\left(b+13\right)}{2b-13}\)
\(=\frac{2b+39}{2b+39}-\frac{2b-13}{2b-13}=1-1=0\)
Ta có:\(\frac{3a-b}{2a+15}=\frac{3a-b}{2a+a-b}=\frac{3a-b}{3a-b}=1\)
\(\frac{3b-a}{2b-15}=\frac{3b-a}{2b-\left(a-b\right)}=\frac{3b-a}{3b-a}=1\)
=>P=1+1=2
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: +) \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)
+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)
+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)
Có: \(\frac{a}{b}=\frac{5}{6}=>\frac{a}{5}=\frac{b}{6}\)
Đặt \(\frac{a}{5}=\frac{b}{6}=k=>\hept{\begin{cases}a=5k\\b=6k\end{cases}}\)
Thay vào ta có:
A=\(\frac{3.5k-2.6k}{2.5k-3.5k}=\frac{15k-12k}{10k-15k}=\frac{3k}{-5k}=\frac{-3}{5}\)
=> \(A=\frac{-3}{5}\)
Áp dụng dãy tỉ số bằng nhau ta có:
\(\frac{2a+b}{c}\)=\(\frac{2b+c}{a}\)=\(\frac{2c+a}{b}\)=\(\frac{2a+b+2b+c+2c+a}{a+b+c}=\frac{3a+3b+3c}{a+b+c}=3\)
=> \(\frac{2a+b}{c}\)=3
\(\frac{a}{2b+c}=\frac{1}{3}\)
\(\frac{b}{2c+a}=\frac{1}{3}\Rightarrow\frac{3b}{2c+a}=1\)
=> \(A=3+\frac{1}{3}+1=\frac{13}{3}\)
Áp dụng tính chất của dãy tỉ số bằng nhau
\(\Rightarrow\frac{2a+b}{c}=\frac{2b+c}{a}=\frac{2c+a}{b}=\frac{3a+3b+3c}{a+b+c}\)\(=\frac{3\left(a+b+c\right)}{a+b+c}\)\(=3\)
=> \(\hept{\begin{cases}\frac{2a+b}{c}=3\\\frac{2b+c}{a}=3\\\frac{2c+a}{b}=3\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2a+b=3c\\2b+c=3a\\2c+a=3b\end{cases}}\)
\(\Rightarrow A\)\(=\frac{3c}{c}+\frac{a}{3a}+\frac{3b}{3b}=3+\frac{1}{3}+1=\frac{13}{3}\)
\(A=\frac{13}{3}\)
Có: \(\frac{3a+b+2c}{2a+c}=\frac{a+3b+c}{2b}=\frac{a+2b+2c}{b+c}\)
\(\Rightarrow\frac{a+b+c+2a+c}{2a+c}=\frac{a+b+c+2b}{2b}=\frac{a+b+c+b+c}{b+c}\)
\(\Rightarrow\frac{a+b+c}{2a+c}+1=\frac{a+b+c}{2b}+1=\frac{a+b+c}{b+c}+1\)
\(\Rightarrow\frac{a+b+c}{2a+c}=\frac{a+b+c}{2b}=\frac{a+b+c}{b+c}\)
\(\Rightarrow2a+c=2b=b+c\)
\(\Rightarrow\hept{\begin{cases}c=b\\a=\frac{1}{2}b\end{cases}}\)
Thay vào biểu thức trên , ta được:
\(P=\)\(\frac{\left(\frac{1}{2}b+b\right)\left(b+b\right)\left(b+\frac{1}{2}b\right)}{\frac{1}{2}b.b.b}=9\)
Vậy \(P=9\)