2. Ta có:
\(\dfrac{\left(a-b\right)\left(c-d\right)}{\left(b-c\right)\left(d-a\right)}=\dfrac{2023}{2024}\)
\(\Rightarrow\dfrac{\left(b-c\right)\left(d-a\right)}{\left(a-b\right)\left(c-d\right)}=\dfrac{2024}{2023}\)
\(\Rightarrow1-\dfrac{\left(b-c\right)\left(d-a\right)}{\left(a-b\right)\left(c-d\right)}=1-\dfrac{2024}{2023}\)
\(\Rightarrow\dfrac{\left(a-b\right)\left(c-d\right)-\left(b-c\right)\left(d-a\right)}{\left(a-b\right)\left(c-d\right)}=\dfrac{2023-2024}{2023}\)
\(\Rightarrow\dfrac{ac-ad-bc+bd-bd+ab+cd-ac}{\left(a-b\right)\left(c-d\right)}=\dfrac{-1}{2023}\)
\(\Rightarrow\dfrac{ab-ad-bc+cd}{\left(a-b\right)\left(c-d\right)}=-\dfrac{1}{2023}\)
\(\Rightarrow\dfrac{a\left(b-d\right)-c\left(b-d\right)}{\left(a-b\right)\left(c-d\right)}=-\dfrac{1}{2023}\)
\(\Rightarrow\dfrac{\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(c-d\right)}=-\dfrac{1}{2023}\)
Vậy: ....
