\(a+\dfrac{1}{b}=\dfrac{a}{b}\Leftrightarrow\dfrac{ab+1}{b}=\dfrac{a}{b}\Leftrightarrow ab+1=a\left(1\right)\)
\(\dfrac{a}{b}=-4\Leftrightarrow a=-4b\left(2\right)\)
Thay (2) vào (1), ta được:
\(-4b^2+1=-4b\)
\(\Rightarrow-4b^2+4b+1=0\)
\(\Rightarrow-4\left(b^2+b-\dfrac{1}{4}\right)=0\)
\(\Rightarrow-4\left(b^2+2\cdot b\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+2=0\)
\(\Rightarrow-4\left(b+\dfrac{1}{2}\right)^2=-2\)
\(\Rightarrow\left(b+\dfrac{1}{2}\right)^2=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}b+\dfrac{1}{2}=\sqrt{\dfrac{1}{2}}\\b+\dfrac{1}{2}=-\sqrt{\dfrac{1}{2}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}b=\dfrac{-1+\sqrt{2}}{2}\\b=\dfrac{-1-\sqrt{2}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}b=\dfrac{-1+\sqrt{2}}{2}\\a=2-2\sqrt{2}\end{matrix}\right.\\\left\{{}\begin{matrix}b=\dfrac{-1-\sqrt{2}}{2}\\a=2+2\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy ..................................
