Được 2 cách chứ!
Cách 1:
\(a+b=c+d\Rightarrow a=c+d-b\)
Mà \(ab+1=cd\)
Do đó: \(\left(c+d-b\right)b+1=cd\)
\(\Leftrightarrow bc+b\left(d-b\right)+1=cd\)
\(\Leftrightarrow cd-bc-b\left(d-b\right)=1\)
\(\Leftrightarrow c\left(d-b\right)-b\left(d-b\right)=1\)
\(\Leftrightarrow\left(d-b\right)\left(c-b\right)=1\)
\(\Leftrightarrow\left[{}\begin{matrix}d-b=c-b=1\\d-b=c-b=-1\end{matrix}\right.\)
\(\Rightarrow c=d\)
Cách 2:
\(a+b=c+d\Rightarrow\left\{{}\begin{matrix}b\left(a+b\right)=b\left(c+d\right)\\ab+b^2=bc+bd\end{matrix}\right.\)
Mà \(ab+1=cd\)
Do đó: \(\left(ab+b^2\right)-\left(ab+1\right)=\left(bc+bd\right)-cd\)
\(\Leftrightarrow ab+b^2-ab-1=bc+bd-cd\)
\(\Leftrightarrow b^2-bc-bd+cd=1\)
\(\Leftrightarrow b\left(b-c\right)-d\left(b-c\right)=1\)
\(\Leftrightarrow\left(b-c\right)\left(b-d\right)=1\)
\(\Leftrightarrow\left[{}\begin{matrix}b-c=b-d=1\\b-c=b-d=1\end{matrix}\right.\)
\(\Rightarrow c=d\)
