tìm số tự nhiên a và b biết ƯCLN(a,b)=8 ;a+b=48
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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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ta có \(UCLN\left(a,b\right)\le a,b\)\(\Rightarrow UCLN\left(a,b\right)\le a+b\) điều này mâu thuẫn với giả thiết
\(\hept{\begin{cases}a+b=8\\UCLN\left(a,b\right)=9\end{cases}}\) vậy không tồn tại hai số a,b thỏa mãn
b. ta có \(UCLN\left(a,b\right)=6\Rightarrow\hept{\begin{cases}a=6k\\b=6h\end{cases}}\)với h,k nguyên tố cùng nhau
\(a.b=36h.k=720\Leftrightarrow hk=20=1.2^2.5\) nên \(\left(h,k\right)=\left(1,20\right)\text{ hoặc (4,5)}\)
vậy tương ứng ta có hai bộ số là 6,120 và 24,30 thỏa mãn đề bài
còn a>b nữa
\(Tacó:\)
\(a=8a`;b=8b`\Rightarrow a+b=8\left(a`+b`\right)\Rightarrow a`+b`=6\)
\(\left(a< b\right)\Rightarrow a`< b`\left(a`,b`\ne0\right)\)
\(\Rightarrow b`\in\left\{4;5\right\}\)
\(+b`=4\Rightarrow b=32\Rightarrow a=16\Rightarrow UCLN\left(a,b\right)=16\left(loại\right)\)
\(+b`=5\Rightarrow b=40\Rightarrow a=8\Rightarrow UCLN\left(a,b\right)=8\left(thoảman\right)\)
Vậy: a=8;b=40