a/Ta có :
\(ƯCLN\left(a,b\right)=13\) \(\Leftrightarrow\left\{{}\begin{matrix}a=13.a_1\\b=13.b_1\\ƯCLN\left(a_1;b_1\right)=1\end{matrix}\right.\)
Mà \(a+b=117\)
\(\Leftrightarrow13a_1+13b_1=117\)
\(\Leftrightarrow13\left(a_1+b_1\right)=117\)
\(\Leftrightarrow a_1+b_1=9\)
Ta có các trường hợp :
+) \(\left\{{}\begin{matrix}a_1=1\\b_1=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=13\\b=104\end{matrix}\right.\)
+) \(\left\{{}\begin{matrix}a_1=2\\b_1=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=26\\b=91\end{matrix}\right.\)
+) \(\left\{{}\begin{matrix}a_1=4\\b_1=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=52\\b=65\end{matrix}\right.\)
Vậy....
b/ Ta có :
\(BCNN\left(a.b\right).ƯCLN\left(a,b\right)=a.b=2400\)
\(\LeftrightarrowƯCLN\left(a,b\right)=2400:120=20\)
\(ƯCLN\left(a,b\right)=20\) \(\Leftrightarrow\left\{{}\begin{matrix}a=20a_1\\b=20b_1\\ƯCLN\left(a_1,b_1\right)=1\end{matrix}\right.\)
Mà \(a.b=2400\)
\(\Leftrightarrow20.a_1.20.b_1=2400\)
\(\Leftrightarrow a_1.b_1=6\)
Ta có các trường hợp :
+) \(\left\{{}\begin{matrix}a_1=1\\b_1=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=20\\b=120\end{matrix}\right.\)
+) \(\left\{{}\begin{matrix}a_1=2\\b_1=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=40\\b=60\end{matrix}\right.\)
+) \(\left\{{}\begin{matrix}a_1=3\\b_1=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=60\\b=40\end{matrix}\right.\)
+) \(\left\{{}\begin{matrix}a_1=6\\b_1=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=120\\b=20\end{matrix}\right.\)
Vậy..
1) Ta có: a=13m; b=13n (m;n)=1
Thay vào: a+b= 13m+13n=13(m+n)=117
m+n=117:13= 9
(m;n)= (1;8) (2;7) (4;5) (5;4) (7;2) (8;1)
\(\Rightarrow\) (a;b)= ( 13; 104) (26;91) (52;65) (65;52) (91;26) (104;13)
