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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: 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}
\(a,40=2^3.5\\ 24=2^3.3\\ \Rightarrow UCLN_{\left(24;40\right)}=2^3=8\\ \RightarrowƯC_{\left(40;24\right)}=Ư_{\left(8\right)}=\left\{1;2;4;8\right\}\\ b,80=2^4.5\\ 144=2^4.3^2\\ \Rightarrow UCLN_{\left(80;144\right)}=2^4=16\\ \RightarrowƯC_{\left(80;144\right)}=Ư_{\left(16\right)}=\left\{1;2;4;8;16\right\}\\ c,9=3^2\\ 18=2.3^2\\ 72=3^2.2^3\\ \Rightarrow UCLN_{\left(8;18;72\right)}=3^2=9\\ \RightarrowƯC_{\left(9;18;72\right)}=Ư_{\left(9\right)}=\left\{1;3;9\right\}\\ d,25=5^2\\ 55=5.11\\ 75=5^2.3\\ \Rightarrow UCLN_{\left(25;55;75\right)}=5\\ \RightarrowƯC_{\left(25;55;75\right)}=Ư_{\left(5\right)}=\left\{1;5\right\}\)
\(a,ƯC\left(40,24\right)=Ư\left(8\right)=\left\{...\right\}\\ b,ƯC\left(12,52\right)=Ư\left(4\right)=\left\{...\right\}\\ c,ƯC\left(36,990\right)=Ư\left(18\right)=\left\{...\right\}\\ d,ƯC\left(54,36\right)=Ư\left(9\right)=\left\{...\right\}\\ e,ƯC\left(10,20,70\right)=Ư\left(10\right)=\left\{...\right\}\\ f,ƯC\left(25,55,75\right)=Ư\left(5\right)=\left\{...\right\}\\ g,ƯC\left(80,144\right)=Ư\left(16\right)=\left\{...\right\}\\ h,ƯC\left(63,2970\right)=Ư\left(9\right)=\left\{...\right\}\\ i,ƯC\left(65,125\right)=Ư\left(5\right)=\left\{...\right\}\\ j,ƯC\left(9,18,72\right)=Ư\left(9\right)=\left\{...\right\}\\ k,ƯC\left(24,36,60\right)=Ư\left(12\right)=\left\{...\right\}\\ l,ƯC\left(16,42,86\right)=Ư\left(2\right)=\left\{..\right\}\)
Bài 1:
\(101\cdot125+101\cdot25-101\cdot50\)
\(=101\cdot\left(125+25-50\right)\)
\(=101\cdot100\)
\(=10100\)
Bài 2:
\(76\cdot115+56\cdot24+59\cdot24\)
\(=76\cdot115+24\cdot\left(56+59\right)\)
\(=76\cdot115+24\cdot115\)
\(=115\cdot\left(76+24\right)\)
\(=115\cdot100\)
\(=11500\)
a. -15 . 999
=-14985
b. 121. (-63) +63 .(-53) -63.26
=-7623+(-3339)-1638
=-10962-1638
=-12600
c. [299. (-74) + (-299) . (-24)] : (-50).
=[-22126+7176]:(-50)
=-14950:(-50)
=299
d. [900 + (-1140) + 720 ] : (-120)
=[-240+720]:(-120)
=480:(-120)
=-4
e. 6. (-4 2 ). (-102 ) : 24
=-252.(-102):24
=25704:24
=1071
f. [(-8).(-8).(-8).(-8) + 84 ] : 8100
=[64.(-8).(-8)+84]:8100
=[-512.(-8)+84]:8100
=[4096+84]:8100
=4180:8100
=0,51
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:
\(50=2.5^2\)
\(85=5.17\)
\(70=2.5.7\)
\(\Rightarrow UCLN\left(50,70,85\right)=5^{ }\)