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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: +) \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)
+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)
+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)
Em kiểm tra lại đề ở tỉ số đầu tiên
\(\dfrac{2a+2b-2c}{c}=\dfrac{2b-2c+2a}{a}\)
Hay là: \(\dfrac{2a+2b-2c}{c}=\dfrac{2b+2c-2a}{a}\)
Vì \(a,b,c>0\Rightarrow a+b+c\ne0\)
Áp dụng tc dtsbn:
\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\\ \Rightarrow\left\{{}\begin{matrix}2b+c-a=2a\\2c-b+a=2b\\2a+b-c=2c\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a-2b=c\\3b-2c=a\\3c-2a=b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a-c=2b\\3b-a=2c\\3c-b=2a\end{matrix}\right.\\ \Rightarrow P=\dfrac{abc}{2a\cdot2b\cdot2c}=\dfrac{1}{8}\)
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 \(P=9\)
ta có : 3/a+b=2/b+c=1/c+a=>a+b/3=b+c/2=c+a/1
=a+b-b-c/3-2=a-c/1
=>c+a=a-c=>c=0=>b=2a
thay c=0;b=2a vào M ta đc:
M=2a+3.2a+2020.0/3a+2.2a-2021.0=8a/7a=8/7
