Vì \(\dfrac{x}{y}=2\Rightarrow x=2y\)

Ta có: \(\dfrac{2x-y}{x+2y}=\dfrac{2.2y-y}{2y+2y}=\dfrac{4y-y}{4y}=\dfrac{3y}{4y}=\dfrac{3}{4}\)

Để \(A\in Z\Rightarrow4n-1⋮2n+3\)

Ta có:\(\left\{{}\begin{matrix}4n-1⋮2n+3\\2n+3⋮2n+3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4n-1⋮2n+3\\2\left(2n+3\right)⋮2n+3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4n-1⋮2n+3\\4n+6⋮n+3\end{matrix}\right.\)

\(\Rightarrow\left(4n+6\right)-\left(4n-1\right)⋮2n+3\)

\(\Rightarrow4n+6-4n+1⋮2n+3\Rightarrow4n-4n+6-1⋮2n+3\)

\(\Rightarrow5⋮2n+3\Rightarrow2n+3\inƯ\left(5\right)\)

\(\Rightarrow2n+3\in\left\{\pm1;\pm5\right\}\)

\(\Rightarrow2n\in\left\{-8;-4;\pm2\right\}\)

\(\Rightarrow n\in\left\{-4;-2;\pm1\right\}\)

Vậy \(n\in\left\{-4;-2;\pm1\right\}\)

mk nghĩ câu b sai đề thì phải

a, Ta có: \(\overline{abcdeg}=\overline{ab}.10000+\overline{cd}.100+\overline{eg}=\overline{ab}.9999+\overline{ab}+\overline{cd}.99+\overline{cd}+\overline{eg}\)

\(=\overline{ab}.9999+\overline{cd}.99+\left(\overline{ab}+\overline{cd}+\overline{eg}\right)\)

Vì : \(\left\{{}\begin{matrix}9999⋮11;99⋮11\Rightarrow\overline{ab}.9999⋮11;\overline{cd}.99⋮11\Rightarrow\overline{ab}.9999+\overline{cd}.99⋮11\\\overline{ab}+\overline{cd}+\overline{eg}⋮11\end{matrix}\right.\)

Nên \(\overline{abcdeg}⋮11\)

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