Giải:
Gọi \(ƯCLN\left(a^2;a+b\right)=1\)
\(\Rightarrow\left\{\begin{matrix}a^2⋮d\\a+b⋮d\end{matrix}\right.\Rightarrow\left\{\begin{matrix}a\left(a+b\right)⋮d\\a^2+ab⋮d\end{matrix}\right.\)
\(\Rightarrow a^2+ab-a^2⋮d\)
\(\Rightarrow ab⋮d\)
Mà \(\left(a;b\right)=1\Rightarrow\left\{\begin{matrix}a⋮d\\b⋮d\end{matrix}\right.\)
Nếu \(a⋮d\)
\(\Rightarrow a+b⋮d\Rightarrow b⋮d\)
\(\Rightarrow d\inƯC\left(a;b\right)\)
Mà \(ƯCLN\left(a;b\right)=1\Rightarrow d=1\RightarrowƯCLN\left(a^2;a+b\right)=1\)
Nếu \(b⋮d\)
\(\Rightarrow a+b⋮d\Rightarrow a⋮d\)
\(\Rightarrow d\inƯC\left(a;b\right)\)
Mà \(ƯCLN\left(a;b\right)=1\Rightarrow d=1\RightarrowƯCLN\left(a^2;a+b\right)=1\)
Vậy nếu \(\left(a;b\right)=1\) thì \(\left(a^2;a+b\right)=1\) (Đpcm)
