Question

Tính \(A=\frac{2019x+2020y}{2019x-2020y}\) biết \(\frac{x}{2}=\frac{y}{3}\)

Answer 1

Ta có:

\(\frac{x}{2}=\frac{y}{3}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)

\(\Rightarrow A=\frac{2019x+2020y}{2019x-2020y}=\frac{2019.2k+2020.3k}{2019.2k-2020.3k}=\frac{10098k}{-2022k}=\frac{10098}{-2022}=\frac{-1683}{337}\)

Answer 2

Ta có:

\(\frac{x}{2}=\frac{y}{3}.\)

Đặt \(\frac{x}{2}=\frac{y}{3}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\)

Lại có: \(A=\frac{2019x+2020y}{2019x-2020y}.\)

+ Thay \(x=2k\) và \(y=3k\) vào A ta được:

\(A=\frac{2019.2k+2020.3k}{2019.2k-2020.3k}\)

\(\Rightarrow A=\frac{4038k+6060k}{4038k-6060k}\)

\(\Rightarrow A=\frac{k.\left(4038+6060\right)}{k.\left(4038-6060\right)}\)

\(\Rightarrow A=\frac{4038+6060}{4038-6060}\)

\(\Rightarrow A=\frac{10098}{-2022}\)

\(\Rightarrow A=\frac{-1683}{337}.\)

Vậy \(A=\frac{-1683}{337}.\)

Chúc bạn học tốt!