1: Do (a, b) = 19 nên tồn tại x, y sao cho (x, y) = 1 và \(\left\{{}\begin{matrix}a=19x\\b=19y\end{matrix}\right.\).
Suy ra \(95=a+b=19x+19y\Rightarrow x+y=5\).
Mặt khác, do (x, y) = 1 nên \(\left(x,y\right)\in\left\{\left(1;4\right),\left(2;3\right),\left(3;2\right),\left(4;1\right)\right\}\).
Suy ra \(\left(a,b\right)\in\left\{\left(19;76\right),\left(38;57\right),\left(57;38\right),\left(76;19\right)\right\}\).
Giải:
1.Do (a, b) = 19 nên tồn tại x, y sao cho (x, y) = 1 và \(\left\{{}\begin{matrix}a=19x\\b=19y\end{matrix}\right.\)
Suy ra 95=a+b=19x+19y⇒x+y=5
Mặt khác, do (x, y) = 1 nên (x,y) ∈ {(1;4),(2;3),(3;2),(4;1)}
Suy ra (a,b) ∈ {(19;76),(38;57),(57;38),(76;19)}
2. Ta có:
\(ƯCLN\left(a;b\right).BCNN\left(a;b\right)=a.b\)
\(\RightarrowƯCLN\left(a;b\right)=2400:120=20\)
\(\Rightarrow\left\{{}\begin{matrix}a=20.m\\b=20.n\end{matrix}\right.\) với m;n ∈ N; (m;n)=1
Ta có:
a.b=2400
20.m.20.n=2400
400.m.n=2400
m.n=6
\(\Rightarrow\left(m;n\right)\inƯ\left(6\right)=\left(1;2;3;6\right)\)
Ta có bảng giá trị:
m=1 thì n=6 ➜a=20;b=120
m=6 thì n=1 ➜a=120;b=20
m=2 thì n=3 ➜a=40;b=60
m=3 thì n=2 ➜a=60;b=20
\(\Rightarrow\left(a;b\right)=\left\{\left(20;120\right);\left(120;20\right);\left(40;60\right)\left(60;40\right)\right\}\)
3.
Ta có:
\(\Rightarrow\left\{{}\begin{matrix}a=2.m\\b=2.n\end{matrix}\right.\) với m;n ∈ N; (m;n)=1
Ta có:
a.b=96
2.m.2.n=96
4.m.n=96
m.n=24
\(\Rightarrow\left(m;n\right)\inƯ\left(24\right)=\left(1;2;3;4;6;8;12;24\right)\)
\(\Rightarrow\left(a;b\right)=\left\{\left(2;48\right);\left(48;2\right);\left(4;24\right);\left(24;4\right);\left(6;16\right);\left(16;6\right);\left(8;12\right);\left(12;8\right)\right\}\) 4. Ta có:
\(ƯCLN\left(a;b\right).BCNN\left(a;b\right)=a.b\)
\(\Rightarrow a.b=15.1260=18900\)
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