Bài 5
Gọi d là ƯC ( 12n + 1 ; 30n + 2 )
⇒ \(\left\{{}\begin{matrix}\text{12n + 1 ⋮ d}\\30n+2⋮d\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}\text{5.( 12n + 1 ) ⋮ d }\\2.(30n+2)⋮d\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}\text{60n + 5 ⋮ d }\\60n+4⋮d\end{matrix}\right.\)
⇒ [ ( 60n + 5 ) - ( 60n + 4 ) ] ⋮ d
⇒ 1 ⋮ d ⇒ d = 1
Vậy \(\dfrac{12n+1}{\text{30n + 2}}\) là phân số tối giản
\(a,\widehat{yOz}=\widehat{xOz}-\widehat{yOx}=120^0-40^0=80^0\\ b,\widehat{xOy}+\widehat{xOt}=180^0\left(kề.bù\right)\\ \Rightarrow\widehat{xOt}=180^0-40^0=140^0\\ c,Om.là.p/g.\widehat{yOz}.nên.\widehat{mOy}=\dfrac{1}{2}\widehat{yOz}=\dfrac{1}{2}\cdot80^0=40^0\\ \Rightarrow\widehat{mOy}=\widehat{xOy}\left(=40^0\right)\)
Vậy Oy là p/g \(\widehat{xOm}\)
