Ta có:
90 = 2 x 32 x 5
126 = 2 x 32 x 7
=> ƯCLN(90; 126) = 2 x 32 = 18
=> ƯC(90; 126) = Ư(18) = {1; 2; 3; 6; 9; 18}
\(126=2.3^2.7\)
=> UCLN[ 90;126]=\(2.3^2=18\)
=> UC[90;126] = U[18] ={1;2;3;6;9;18}
=> Uc [ 90 ;126 ] ={1;2;3;6;9;18}
TA CÓ : 90 = 2 . 32 . 5
126 = 2. 32. 7
=> ƯCLN (90, 126) = 2 . 32 = 18
Vì ƯC(90,126) = Ư(18)
Mà Ư(18) = { -18; - 9; -6; -3; - 2; -1; 1; 2; 3 ; 6; 9; 18}
=> ƯC(90,126) \(\in\) { -18; -9; -6; -3; -2; -1; 1;2;3;6;9;18}
Ta có : \(90=2.3^2.5\)
\(126=2.3^2.7\)
\(\Rightarrow UwCLN\left(90;126\right)=2.3^2=18\)
\(\RightarrowƯC\left(90;126\right)=Ư\left(18\right)=\left\{1;2;3;6;9;18\right\}\)
\(\RightarrowƯC\left(90;126\right)=\left\{1;2;3;6;9;18\right\}\)
Bài giải :
Ta có :
90 = 2 . 32 . 5
126 = 2 . 32 . 7
ƯCLN(90 , 126) = 2 . 32 = 18.
Ư(18) = { 1 ; 2 ; 3 ; 6 ; 9 ; 18 }
ƯC(90 , 126 ) = { 1 ; 2 ; 3 ; 6 ; 9 ; 18 }