24 ; 40 ; 36

24 = 2³.3

40 = 2³.5

36 = 2².3²

⇒ ƯCLN ( 24 , 40 , 36 ) = 2² = 4 

Ư(4) = { 1 , 2 , 4 }

⇒ ƯC ( 24 , 40 , 36 ) = { 1 2 ,3 }

40 = 2³.5

48 = 2⁴.3

ƯCLN(40; 48) = 2³ = 8

Ta có:

\(40=2^3\cdot5\\48=2^4\cdot3\\\Rightarrow \mathit{UCLN}(40;48)=2^3=8\)

Phân tích ra thừa số nguyên tố$:$

$40=5.2^{3}$

$48=2^{4}.3$

$ƯCLN(40;48)=2^{3}=8$

ƯCLN=12

ƯCLN của 48,36,24=12 nha

Ta có: 

\(28=2^2\cdot7\\48=2^4\cdot3\)

\(\Rightarrow\left\{{}\begin{matrix}UCLN\left(28;48\right)=2^2=4\\BCNN\left(28;48\right)=2^4\cdot3\cdot7=336\end{matrix}\right.\)

Ta có:

\(28=2^2\cdot7\)

\(48=2^4\cdot3\)

\(\RightarrowƯCLN\left(28;48\right)=2^2=4\)

\(\Rightarrow BCNN\left(28;48\right)=2^4\cdot3\cdot7=336\)

8

ƯCLN(32; 40) = 2³ = 8

=8

   40 = 23.5

   60 = 22.3.5

   ƯCLN(40;60) = 22.5 = 20

Ta có:

\(40=2^3.5\)

\(56=2^3.7\)

\(\RightarrowƯCLN\left(40;56\right)=2^3=8\)

ƯCLN(40;56)=23=8

+)40=2^3.5
56=2^3.7
+)BCNN(40,56)=2^3.5.7=280

:)))

\(16=2^4;28=2^2\cdot7;40=2^3\cdot5\\ \RightarrowƯCLN\left(16,28,40\right)=2^2=4\\ BCNN\left(16,28,40\right)=2^4\cdot5\cdot7=560\)

16=2⁴ ; 28=2².7 ; 40=2³.5

=> ước chung lớn nhất: 2² 

     bội chung nhỏ nhất: 2².7.5=140

16 bạn nhé

16 for sure