Question

Cho a,b thoa man 3a > b > 0 va 3a² + 2b² - 7ab = 0. Tinh P = \(\frac{2019a-2020b}{2020a+2021b}\)

Answer 1

\(3a^2+2b^2-7ab=0\)

\(\Leftrightarrow\left(3a^2-6ab\right)+\left(2b^2-ab\right)=0\)

\(\Leftrightarrow3a\left(a-2b\right)-b\left(a-2b\right)=0\)

\(\Leftrightarrow\left(3a-b\right)\left(a-2b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3a-b=0\\a-2b=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}b=3a\\a=2b\end{matrix}\right.\)

Thay \(b=3a\) vào P ta có :

\(P=\frac{2019a-2020.3a}{2020a+2021.3a}=\frac{-3951a}{8083a}=\frac{-3951}{8083}\)

Thay \(a=2b\) vào P ta có :

\(P=\frac{2019.2b-2020b}{2020.2b+2021b}=\frac{2018}{6061}\)

Vậy..