Question

[Nguyễn Thanh Liêm]

Cho dãy tỉ số \(\frac{3a-2b}{5}=\frac{2c-5a}{3}=\frac{5b-3c}{2}\) .tính các giá trị của biểu thức P\(=\frac{2a+3b+4c}{3a+4b+5c}\)

nhờ mọi mina đó

Answer 1

[Y]

\(\frac{3a-2b}{5}=\frac{2c-5a}{3}=\frac{5b-3c}{2}=\frac{15a-10b}{25}=\frac{6c-15a}{9}=\frac{10b-6c}{4}\)

\(=\frac{15a-10b+6c-15a+10b-6c}{25+9+4}=0\)

\(\Rightarrow\left\{{}\begin{matrix}3a=2b\\2c=5a\\5b=3c\end{matrix}\right.\Rightarrow\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=\frac{a+b+c}{10}\)

\(\Rightarrow\left\{{}\begin{matrix}a=\frac{a+b+c}{5}\\b=\frac{3\left(a+b+c\right)}{10}\\c=\frac{a+b+c}{2}\end{matrix}\right.\)

\(\Rightarrow P=\frac{\frac{33\left(a+b+c\right)}{10}}{\frac{43\left(a+b+c\right)}{10}}=\frac{33}{43}\)