Ta có: (a, b) = 45 \(\Leftrightarrow\left\{{}\begin{matrix}a=45a'\\b=45b'\\\left(a',b'\right)=1\end{matrix}\right.\)
Ta lại có: [a, b] = \(\dfrac{ab}{\left(a,b\right)}=\dfrac{45^2a'b'}{45}=45a'b'\)
Ta lại có: (a, b) + [a, b] = 270
\(\Rightarrow45a'b'+45=270\)
\(\Rightarrow a'b'+1=6\)
\(\Rightarrow a'b'=5\)
Mà a > b nên a' > b' \(\Rightarrow\left\{{}\begin{matrix}a'=5\\b'=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=225\\b=45\end{matrix}\right.\)
Ta có: \(\left(a,b\right)=5\Leftrightarrow\left\{{}\begin{matrix}a=5a'\\b=5b'\\\left(a',b'\right)=1\end{matrix}\right.\)
Ta lại có: \(ab=300\)
\(\Rightarrow25a'b'=300\)
\(\Rightarrow a'b'=12\)
Mà a > b nên a' > b' \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a'=12\\b'=1\end{matrix}\right.\\\left\{{}\begin{matrix}a'=4\\b'=3\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=60\\b=5\end{matrix}\right.\\\left\{{}\begin{matrix}a=20\\b=15\end{matrix}\right.\end{matrix}\right.\)
