Ta có :
\(\dfrac{a}{b}=\dfrac{60}{180}=\dfrac{5}{9}\) \(\left(a,b\in N\right)\)
Gọi \(ƯCLN\left(a,b\right)=d\Rightarrow\left\{{}\begin{matrix}a=d.a_1\\b=d.b_1\\ƯCLN\left(a_1;b_1\right)=1\end{matrix}\right.\)
\(\Rightarrow BCNN\left(a,b\right)=d.a_1.b_1=180\)\(\left(1\right)\)
Mà \(\dfrac{a}{b}=\dfrac{5}{9}\Rightarrow\dfrac{d.a_1}{d.b_1}=\dfrac{5}{9}\)
\(\Rightarrow\dfrac{a_1}{b_1}=\dfrac{5}{9}\) \(\Rightarrow\left\{{}\begin{matrix}a_1=5\\b_1=9\end{matrix}\right.\) (do \(ƯCLN\left(a_1;b_1\right)=1\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow d.5.9=180\)
\(\Rightarrow d=4\)
\(\Rightarrow\left\{{}\begin{matrix}a=4.5=20\\b=4.9=36\end{matrix}\right.\) (thỏa mãn \(a,b\in Z;b\ne0\))
Vậy phân số \(\dfrac{a}{b}\) cần tìm là \(\dfrac{20}{36}\)
~ Chúc bn học tốt ~
Giải:
Ta có: \(\dfrac{a}{b}=\dfrac{60}{108}\Rightarrow\dfrac{a}{b}=\dfrac{5}{9}\Rightarrow\dfrac{a}{5}=\dfrac{b}{9}\)
Đặt \(\dfrac{a}{5}=\dfrac{b}{9}=k\Rightarrow\left\{{}\begin{matrix}a=5k\\b=9k\end{matrix}\right.\)
Từ đó \(BCNN\left(a,b\right)=BCNN\left(5k,9k\right)=45k=180\)
\(\Rightarrow k=4\)
\(\Rightarrow\left\{{}\begin{matrix}a=20\\b=36\end{matrix}\right.\)
Vậy a = 20, b = 36
