Ta có :
\(ƯCLN\left(a,b\right)=56\) \(\Leftrightarrow\left\{{}\begin{matrix}a=56k\\b=56k_1\end{matrix}\right.\) \(\left(ƯCLN\left(k,k_1\right)=1\right)\) \(\left(1\right)\)
Thay \(\left(1\right)\) vào \(a+b=224\) ta được :
\(56k+56k_1=224\)
\(\Leftrightarrow56\left(k+k_1\right)=224\)
\(\Leftrightarrow k+k_1=4\)
Mà \(\left(k;k_1\right)=1\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}k=1\\k_1=3\end{matrix}\right.\\\left\{{}\begin{matrix}k=3\\k_1=1\end{matrix}\right.\end{matrix}\right.\)
+) \(\left\{{}\begin{matrix}k=1\\k_1=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=56\\b=168\end{matrix}\right.\)
+) \(\left\{{}\begin{matrix}k=3\\k_1=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=168\\b=56\end{matrix}\right.\)
Vậy ..
\(\left(a,b\right)=56\Leftrightarrow\left\{{}\begin{matrix}a=56a'\\b=56b'\\\left(a',b'\right)=1\end{matrix}\right.\)
Ta có:
\(a+b=224\)
\(\Rightarrow56a'+56b'=224\)
\(\Rightarrow a'+b'=4\)
Giả sử a \(\ge\) b thì a' \(\ge\) b'. Mà (a', b') = 1 và a' + b' = 4 nên a' = 3, b' = 1 \(\Rightarrow\) a = 168; b = 56
