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a) ta có UCLN(a;b).BCNN(a;b)=a.b=120.10=1200
UCLN(a;b)=10 \(\Rightarrow\)\(\left\{{}\begin{matrix}a⋮10\\b⋮10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=10k\\b=10h\end{matrix}\right.\left(k;h\right)=1;k\ge h\)
a.b=1200\(\Leftrightarrow\)10k.10h=1200
nên k.h =1200:100=12
mà (k;h)=1 nên (k;h)=(12;1);(4;3)
nên (a;b)=(120;10);(40;30)
a) goi hai so la a ; b va a >b
vi UCLN(a,b)=18=>a=18k ; b=18q (trong do UCLN (k,q)=1 va k>q)
=>a+b=162
18k+18q =162
18(k+q)=162
k+q=9
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21453
52542000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 | 542454550212.100000000000000000000000000000000000000000000000000000000000000000000000000000 |
a) \(24=2^3.3\)
\(60=2^2.3.5\)
\(UCLN\left(a;b\right)=UCLN\left(24;60\right)=2^2.3=6\)
\(BCNN\left(a;b\right)=BCNN\left(24;60\right)=2^3.3.5=120\)
\(a.b=UCLN\left(a;b\right).BCNN\left(a;b\right)\)
\(\Rightarrow a.b=6.120=720\)
mà \(\dfrac{a}{b}=\dfrac{24}{60}\Rightarrow\dfrac{a}{24}=\dfrac{b}{60}=\dfrac{720}{24.60}=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}a=24.\dfrac{1}{2}=12\\b=60.\dfrac{1}{2}=30\end{matrix}\right.\)
Vậy Phân số cần tìm là \(\dfrac{12}{30}\)
b) \(\left\{{}\begin{matrix}14=2.7\\21=3.7\end{matrix}\right.\)
\(\Rightarrow UCLN\left(a;b\right)=UCLN\left(14;21\right)=7\)
\(a.b=UCLN\left(14;21\right).BCNN\left(14;21\right)\)
\(\Rightarrow a.b=7.3456=24192\)
\(\dfrac{a}{b}=\dfrac{14}{21}\Rightarrow\dfrac{a}{14}=\dfrac{b}{21}=\dfrac{a.b}{14.21}=\dfrac{24192}{294}=\dfrac{576}{7}\)
\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{576}{7}.14=1152\\b=\dfrac{576}{7}.21=1728\end{matrix}\right.\)
Vậy phân số cần tìm là \(\dfrac{1152}{1728}\)
Ta có:
\(735=3\cdot5\cdot7^2\)
\(441=3^2\cdot7^2\)
\(294=2\cdot3\cdot7^2\)
\(\RightarrowƯCLN=\left(735,441,294\right)=3\cdot7^2=147\)
c, Gọi ƯCLN(a; b) = d; d \(\in\) k
⇒ d = 1944 : 108 = 18
⇒ a = 18.k; b = 18.n (k;n) =1; k;n \(\in\) N*
⇒18.k.18.n = 1944
⇒k.n =1944 : (18.18)
k.n = 6
6 = 2.3 Ư(6) = {1; 2; 3;6)
⇒(k; n) = (1; 6); (2; 3); (3; 2); (6; 1)
⇒ (a; b) = (18; 108); (36; 54); (54; 36); (108; 18)
Vì a> b nên (a; b) = (54; 36); (108; 18)
a, a + b = 72; Ư CLN(a; b) = 9 (a > b)
a = 9.k; b = 9.d (k; d) = 1; k; d \(\in\) N*; k >d
9.k + 9.d = 72
9.(k + d) = 72
k + d = 72 : 9
k + d = 8
(k; d) =(1; 7); (2; 6); (3; 5); (4; 4); (5; 3); (6; 2); (7; 1)
vì (k;d) = 1; k > d ⇒ (k;d) = (5; 3); (7; 1)
⇒ (a; b) = (45; 27); (63; 9)
24=2.17
40=2^3.5
=>ƯCLN(24;40)=2^3=8
Tống của tất cả các số có 4 chữ số khác nhau được viết bởi 1 ; 2 ; 3 ; 4 là ...