Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

\(\Rightarrow\frac{2019a^2+2020b^2}{2019a^2-2020b^2}=\frac{2019b^2k^2+2020b^2}{2019b^2k^2-2020b^2}\)

\(=\frac{2019k^2+2020}{2019k^2-2020}\)(1)

và\(\Rightarrow\frac{2019c^2+2020d^2}{2019c^2-2020d^2}=\frac{2019d^2k^2+2020d^2}{2019d^2k^2-2020d^2}\)

\(=\frac{2019k^2+2020}{2019k^2-2020}\)(2)

Từ (1) và (2) suy ra \(\frac{2019a^2+2020b^2}{2019a^2-2020b^2}\)\(=\frac{2019c^2+2020d^2}{2019c^2-2020d^2}\left(đpcm\right)\)

Sửa đề: \(\dfrac{2018a-2019b}{2019a+2020b}=\dfrac{2018c-2019d}{2019c+2020d}\)

\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2020a}{2020b}=\dfrac{2020c}{2020d}=\dfrac{2019a}{2019c}=\dfrac{2019b}{2019d}=\dfrac{2018a}{2018c}=\dfrac{2018b}{2018d}=\dfrac{2018a-2019b}{2018c-2019d}=\dfrac{2019a+2020b}{2019c+2020d}\\ \Leftrightarrow\dfrac{2018a-2019b}{2019a+2020b}=\dfrac{2018c-2019d}{2019c+2020d}\)

\(\dfrac{2018a-2019b}{2019c-2020d}=\dfrac{2018c-2018c}{2019a+2020b}\)

Sao .... ;-; ;-; 

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=\dfrac{a+b+c+d}{a+b+c+d}=1\\ \Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=d\\d=a\end{matrix}\right.\Rightarrow a=b=c=d\\ \Rightarrow VT=\left(\dfrac{2019a+2020a-2021a}{2019a+2020a-2021a}\right)^3=1^3=1=\dfrac{a^2}{a\cdot a}=VP\)

a) Áp dụng dãy tỉ số bằng nhau:

 \(\frac{a}{c}=\frac{b}{d}=\frac{2020b}{2020d}=\frac{a+2020b}{c+2020d}=\frac{a-2020b}{c-2020d}\)

=> \(\frac{a+2020b}{c+2020d}=\frac{a-2020b}{c-2020d}\)

=> \(\frac{a+2020b}{a-2020b}=\frac{c+2020d}{c-2020d}\)

b) \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\)

Áp dụng dãy tỉ số bằng nhau:

\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)

=> \(\frac{a}{b}=\frac{a+c}{b+d}\Rightarrow\frac{a}{a+c}=\frac{b}{b+d}\)

=> \(\frac{2020a}{2020\left(a+c\right)}=\frac{b}{b+d}\)

=> \(\frac{2020\left(a+c\right)}{2020a}=\frac{b+d}{b}\)

c) \(2a+3c\left(b+d\right)=\left(a+c\right)\left(2b+3d\right)\)

Câu c sai đề.