Tìm ƯCLN
a) 24; 84; 180
b)50; 70;85
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a: ƯCLN(40;24)=8
ƯC(40;24)=Ư(8)={1;2;4;8}
b: UCLN(80;144)=16
UCLN(80;144)={1;2;4;8;16}
c: UCLN(9;18;72)=9
ƯC(9;18;72)={1;3;9}
d: ƯCLN(25;55;75)=5
ƯC(25;55;75)={1;5}
\(a,40=2^3.5\\ 24=2^3.3\\ \Rightarrow UCLN_{\left(24;40\right)}=2^3=8\\ \RightarrowƯC_{\left(40;24\right)}=Ư_{\left(8\right)}=\left\{1;2;4;8\right\}\\ b,80=2^4.5\\ 144=2^4.3^2\\ \Rightarrow UCLN_{\left(80;144\right)}=2^4=16\\ \RightarrowƯC_{\left(80;144\right)}=Ư_{\left(16\right)}=\left\{1;2;4;8;16\right\}\\ c,9=3^2\\ 18=2.3^2\\ 72=3^2.2^3\\ \Rightarrow UCLN_{\left(8;18;72\right)}=3^2=9\\ \RightarrowƯC_{\left(9;18;72\right)}=Ư_{\left(9\right)}=\left\{1;3;9\right\}\\ d,25=5^2\\ 55=5.11\\ 75=5^2.3\\ \Rightarrow UCLN_{\left(25;55;75\right)}=5\\ \RightarrowƯC_{\left(25;55;75\right)}=Ư_{\left(5\right)}=\left\{1;5\right\}\)
a) ƯCLN ( 16, 24 )
16 = 23 24 = 22.3
ƯCLN ( 16;24 ) = 22= 4
b) ƯCLN ( 60, 90 )
60 = 22.3.5 90=2.32.5
ƯCLN ( 60;90 ) = 2.3.5 = 30
c) ƯCLN ( 24, 84 )
24 = 23.3 84 = 22.3.7
ƯCLN ( 24;84 ) = 22.3 = 12
d) ƯCLN ( 16, 60 )
16 = 24 60 = 22.3.5
ƯCLN ( 16;60 ) = 22=4
e) ƯCLN ( 18, 77 )
18 = 2.32 77=7.11
ƯCLN ( 18; 77 ) = 1
g) ƯCLN ( 18, 90 )
18 = 2.32 90=2.32.5
ƯCLN ( 18;90 ) = 2.32 = 18
h) ƯCLN ( 18, 30, 42 )
18 = 2.32 30 = 2.3.5 42 = 2.3.7
ƯCLN ( 18;30;42 ) = 2.3=6
k) ƯCLN ( 26, 39, 48 )
26 = 2.13 39 = 3.13 48 = 24.3
ƯCLN ( 26;39;48 ) = 1
\(a,ƯC\left(40,24\right)=Ư\left(8\right)=\left\{...\right\}\\ b,ƯC\left(12,52\right)=Ư\left(4\right)=\left\{...\right\}\\ c,ƯC\left(36,990\right)=Ư\left(18\right)=\left\{...\right\}\\ d,ƯC\left(54,36\right)=Ư\left(9\right)=\left\{...\right\}\\ e,ƯC\left(10,20,70\right)=Ư\left(10\right)=\left\{...\right\}\\ f,ƯC\left(25,55,75\right)=Ư\left(5\right)=\left\{...\right\}\\ g,ƯC\left(80,144\right)=Ư\left(16\right)=\left\{...\right\}\\ h,ƯC\left(63,2970\right)=Ư\left(9\right)=\left\{...\right\}\\ i,ƯC\left(65,125\right)=Ư\left(5\right)=\left\{...\right\}\\ j,ƯC\left(9,18,72\right)=Ư\left(9\right)=\left\{...\right\}\\ k,ƯC\left(24,36,60\right)=Ư\left(12\right)=\left\{...\right\}\\ l,ƯC\left(16,42,86\right)=Ư\left(2\right)=\left\{..\right\}\)
Ta có: 247×24+247×66=247×(24+66)
Theo đề bài: y×(24+66)=247×24+247×66
Từ đó ta có: y×(24+66)=247×(24+66)
Suy ra: y=247.
Đáp án C
a: UCLN(24;36)=12
BCNN(24;36)=216
b: BCNN(24;36;60)=360
UCLN(24;36;60)=12
`#3107.101107`
\(48\div24-x=1\\ \Rightarrow2-x=1\\ \Rightarrow x=2-1\\ \Rightarrow x=1\)
Vậy, `x = 1`
____
\(124-2\times\left(x+3\right)=24\\ \Rightarrow2\left(x+3\right)=124-24\\ \Rightarrow2\left(x+3\right)=100\\ \Rightarrow x+3=100\div2\\ \Rightarrow x+3=50\\ \Rightarrow x=50-3\\ \Rightarrow x=47\)
Vậy, `x = 47.`
1) 48 : 24 - x = 1 2) 123 - 2 x (x+3) = 24
2-x=1 2 x (x+3) = 123 - 24
x= 2-1 2 x (x+3) = 99
x= 1 x+3 = 99 : 2
x+3 = \(\dfrac{99}{2}\)
x = \(\dfrac{99}{2}\) -3
x = \(\dfrac{93}{2}\)
a ) Bạn chỉ việc phân tích các số ấy ra thùa số nguyên tố rồi lập các thừa số chung ví dụ.A) UwCLN (24;84;180)=12
Các cau tương tự cậu cứ làm như vậy nha
ta co:
\(50=2.5^2\)
\(85=5.17\)
\(70=2.5.7\)
\(\Rightarrow UCLN\left(50,70,85\right)=5^{ }\)