Ta có : \(10\le ab\le99\Leftrightarrow21\le2ab+1\le201\)
\(2ab+1\) là số chính phương lẻ nên :
\(2ab+1\in\left\{25,49,81,121,169\right\}\)
\(\Leftrightarrow ab\in\left\{12,24,40,60,84\right\}\)
\(\Leftrightarrow3ab+1\in\left\{37,73,121,181,253\right\}\)
\(\Leftrightarrow ab=40\)
Vậy: \(ab=40\)
Ta có : 10≤ab≤99⇔21≤2ab+1≤20110≤ab≤99⇔21≤2ab+1≤201
2ab+12ab+1 là số chính phương lẻ nên :
2ab+1∈{25,49,81,121,169}2ab+1∈{25,49,81,121,169}
⇔ab∈{12,24,40,60,84}⇔ab∈{12,24,40,60,84}
⇔3ab+1∈{37,73,121,181,253}⇔3ab+1∈{37,73,121,181,253}
⇔ab=40⇔ab=40
Vậy: ab=40