\(a=c+d-b\) thế vào đẳng thức sau:
\(\left(c+d-b\right)b+1=cd\Leftrightarrow\left(c-b\right)b+bd+1=cd\)
\(\Leftrightarrow cd-bd-\left(c-b\right)b=1\)
\(\Leftrightarrow d\left(c-b\right)-b\left(c-b\right)=1\)
\(\Leftrightarrow\left(d-b\right)\left(c-b\right)=1\)
\(\Rightarrow d-b=c-b=\pm1\)
\(\Rightarrow\left(d-b\right)-\left(c-b\right)=0\Leftrightarrow d=c\)