11) Tìm ƯCLN,ƯC của:
a)12 và 18
b)24 và 48
c)300 và 280
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1: UCLN(12;18)=6
BCNN(12;18)=36
2: UCLN(24;48)=24
BCNN(24;48)=48
a) Ta có:
\(24=2^3.3\)
\(10=2.5\)
\(ƯCLN\left(24;10\right)=2\)
\(BCNN\left(24;10\right)=2^3.3.5=120\)
__________
b) Ta có:
\(300=2^2.3.5^2\)
\(280=2^3.5.7\)
\(ƯCLN\left(300;280\right)=2^2.5=20\)
\(BCNN\left(300;280\right)=2^3.3.5^2.7=4200\)
__________
c) Ta có:
\(30=2.3.5\)
\(90=2.3^2.5\)
\(ƯCLN\left(30;90\right)=2.3.5=30\)
\(BCNN\left(30;90\right)=2.3^2.5=90\)
_________
d) Ta có:
\(14=2.7\)
\(21=3.7\)
\(56=2^3.7\)
\(ƯCLN\left(14;21;56\right)=7\)
\(BCNN\left(14;21;56\right)=2^3.3.7=168\)
\(#Wendy.Dang\)
3:
a: \(40=2^3\cdot5;24=2^3\cdot3\)
=>\(ƯCLN\left(40;24\right)=2^3=8\)
=>\(ƯC\left(40;24\right)=Ư\left(8\right)=\left\{1;-1;2;-2;4;-4;8;-8\right\}\)
b: \(12=2^2\cdot3;52=2^2\cdot13\)
=>\(ƯCLN\left(12;52\right)=2^2=4\)
=>\(ƯC\left(12;52\right)=\left\{1;-1;2;-2;4;-4\right\}\)
c: \(36=2^2\cdot3^2;990=2\cdot3^2\cdot5\cdot11\)
=>\(ƯCLN\left(36;990\right)=3^2\cdot2=18\)
=>\(ƯC\left(36;990\right)=\left\{1;-1;2;-2;3;-3;6;-6;9;-9;18;-18\right\}\)
2:
a: \(12=2^2\cdot3;18=3^2\cdot2\)
=>\(ƯCLN\left(12;18\right)=2\cdot3=6\)
b: \(12=2^2\cdot3;10=2\cdot5\)
=>\(ƯCLN\left(12;10\right)=2\)
c: \(24=2^3\cdot3;48=2^4\cdot3\)
=>\(ƯCLN\left(24;48\right)=2^3\cdot3=24\)
d: \(300=2^2\cdot3\cdot5^2;280=2^3\cdot5\cdot7\)
=>\(ƯCLN\left(300;280\right)=2^2\cdot5=20\)
TL :
\(40=2^3.5.\)
\(24=2^2.3.2\)
UCLN (40;24) = \(2^3=8\)
BC (8) = \(\left\{1;8;16;24;32;40.....\right\}\)
b) \(48=2^4.3\)
\(120=2^3.5\)
UCLN (48;120 ) = \(2^2.3=12\)
Lời giải:
a.
$12=2^2.3$; $18=2.3^2$
$\Rightarrow ƯCLN(12,18)=2.3=6$
$\Rightarrow ƯC\in Ư(6)\in \left\{1; 2;3;6\right\}$
b.
$48\vdots 24$ nên $ƯCLN(24,48)=24$
$\Rightarrow ƯC(24,48)\in Ư(24)\in \left\{1; 2; 3; 4; 6; 8;12;24\right\}$
c.
$300=2^2.3.5^2$
$280=2^3.5.7$
$\Rightarrow ƯCLN(300,280)=2^2.5=20$
$\Rightarrow ƯC(300,280)\in Ư(20)\in \left\{1;2;4;5;10;20\right\}$