Tìm hai số tự nhiên a, b. Biết ƯCLN(a,b) = 7; ab = 588 và a < b.
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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
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vì ƯCLN(a,b)=6 (a<b)
a=6m
b=6n
với (m,n)=1,m\(\le\)n
a+b=6m+6n=6(m+n)=84
=>m+n=14
m=1 ,n=13,=>a=6,b=78
m=3,n=11,=>a=18,b=66
m=5,n=9,=>a=30,b=54
m=7,n=7,a=42,b=42
bài còn lại cũng tương tự
\(ƯCLN\left(a,b\right)=7\\ \Rightarrow\left\{{}\begin{matrix}a=7p\\b=7q\end{matrix}\right.\left(p< q;p,q\in N\text{*}\right)\\ ab=588\\ \Rightarrow7p\cdot7q=588\\ \Rightarrow pq=12=1\cdot12=2\cdot6=3\cdot4\)
Mà \(p< q\)
\(\left\{{}\begin{matrix}p=1\\q=12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=7\\b=84\end{matrix}\right.;\left\{{}\begin{matrix}p=2\\q=6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=14\\b=42\end{matrix}\right.;\left\{{}\begin{matrix}p=3\\q=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=21\\b=28\end{matrix}\right.\)
Vậy \(\left(a,b\right)\in\left\{\left(7;84\right);\left(14;42\right);\left(21;28\right)\right\}\)
Đáp án: (a,b)={(4,84),(14,42),(21,28)} Giải thích các bước giải: Do Ư C L N ( a , b ) = 7 a, b chia hết cho 7 suy ra a,b là bội của 7 Ta có a b = 588 = 2 2 .3 .7 2 Do Ư C L N ( a , b ) = 7 a, b chia hết cho 7 suy ra a,b là bội của 7 Suy ra tích của a.b tách thành 2 số hạng đều chia hết cho 7 và có a