\(\sqrt[]{a+b}=\dfrac{\overline{ab}}{a+b}=\dfrac{10a+b}{a+b}\left(a;b\in\left\{0;1;2...9\right\};a\ne0\right)\)
\(\Leftrightarrow a+b=\dfrac{\left(10a+b\right)^2}{\left(a+b\right)^2}\)
\(\Leftrightarrow\left(a+b\right)^3=\left(10a+b\right)^2\)
Đặt \(\left\{{}\begin{matrix}x=a+b\\y=10a+b\end{matrix}\right.\)
\(\Leftrightarrow x^3=y^2\)
\(\Rightarrow y=k^3\left(k\in N\right)\)
\(\Rightarrow10a+b=k^3\left(k=2;3;4\right)\) Vì \(10a+b=10\rightarrow99\)
Nếu \(k=2\Rightarrow y=8\Rightarrow\left(a;b\right)=\left(0;8\right);\left(8;0\right)\) không thỏa mãn vì \(a\ne0;\sqrt[]{8+0}\ne\dfrac{80}{8+0}\)
Nếu \(k=3\Rightarrow y=27\Rightarrow\left(a;b\right)=\left(2;7\right);\left(7;2\right)\) thỏa mãn \(\sqrt[]{2+7}=\dfrac{27}{2+7};\sqrt[]{7+2}\ne\dfrac{72}{7+2}\)
Nếu \(\)\(k=4\Rightarrow y=64\left(ktm\right)\)
Vậy \(\overline{ab}=27\) thỏa mãn đề bài
