Ta có :
\(ƯCLN\left(a,b\right)=d\Leftrightarrow\left\{{}\begin{matrix}a=d.a_1\\b=d.b_1\end{matrix}\right.\) \(\left(a_1;b_1\right)=1\)
\(BCNN\left(a,b\right)=\dfrac{ab}{ƯCLN}=\dfrac{ab}{d}\)
Lại có : \(ƯCLN\left(a,b\right)+BCNN\left(a,b\right)=15\)
\(\Leftrightarrow\dfrac{ab}{d}+d=15\)
\(\Leftrightarrow ab+d^2=15d\Leftrightarrow ab=d\left(15-d\right)\)
\(\Leftrightarrow ab⋮d\Leftrightarrow15⋮d\Leftrightarrow d\inƯ\left(15\right)\)
C tự thay vào rồi tìm tiếp nhé
Ta có :
ƯCLN(a,b)=d⇔{a=d.a1b=d.b1ƯCLN(a,b)=d⇔{a=d.a1b=d.b1 (a1;b1)=1(a1;b1)=1
BCNN(a,b)=abƯCLN=abdBCNN(a,b)=abƯCLN=abd
Lại có : ƯCLN(a,b)+BCNN(a,b)=15ƯCLN(a,b)+BCNN(a,b)=15
⇔abd+d=15⇔abd+d=15
⇔ab+d2=15d⇔ab=d(15−d)⇔ab+d2=15d⇔ab=d(15−d)
⇔ab⋮d⇔15⋮d⇔d∈Ư(15)⇔ab⋮d⇔15⋮d⇔d∈Ư(15)
C tự thay vào rồi tìm tiếp nhé
