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Question

Chứng minh rằng: Nếu (a;b) = 1 thì (a²; a + b) =1

Hoang Hung Quan · Accepted Answer

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)

Câu trả lời:

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)