Tìm hai số tự nhiên a và b biết : BCNN(a,b) + ƯCLN(a,b) = 53
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8 tháng 12 2015
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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29 tháng 10 2016
21453
52542000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 | 542454550212.100000000000000000000000000000000000000000000000000000000000000000000000000000 |
(a,b)= ƯCLN(a,b) ; [a,b]= BCNN(a,b)
Gọi d là (a,b)
\(\Rightarrow\hept{\begin{cases}a=dm\\b=dn\end{cases}}\) ( m;n \(\in\) N*) ; ( m;n)= 1
mà a.b = [a,b] . (a,b)
=> dm .dn = [a,b] . d
=> dmn = [a,b]
mà [ a,b ] + (a,b)= 53
dmn + d = 53
d(mn+1) = 53 => d; mn+1 \(\in\) Ư(53)={1;53 }
Ta có bảng sau
vì m.n = 52 mà (m,n)=1 nên ta có bảng
Vậy \(\hept{\begin{cases}a=1\\b=52\end{cases}};\hept{\begin{cases}a=52\\b=1\end{cases};\hept{\begin{cases}a=4\\b=13\end{cases}};\hept{\begin{cases}a=13\\b=4\end{cases}}}\)