a/ \(-7⋮2n+1\)

\(\Leftrightarrow2n+1\inƯ\left(-7\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2n+1=1\\2n+1=7\\2n+1=-1\\2n+1=-7\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n=0\\n=3\\n=-1\\n=-3\end{matrix}\right.\)

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b/ \(-11⋮2n+5\)

\(\Leftrightarrow2n+5\inƯ\left(-11\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2n+5=1\\2n+5=11\\2n+5=-1\\2n+5=-11\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}n=-2\\n=3\\n=-3\\n=-8\end{matrix}\right.\)

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sorry mk ko biet

\(\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2n}\right)=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2n-1}+\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2n}\right)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n-1}+\frac{1}{2n}-\frac{1}{1}-\frac{1}{2}-....-\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\left(\text{đpcm}\right)\)

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nâng cao ak

a) 3n + 5 \(⋮\)2n

\(\Leftrightarrow\)n + 5 \(⋮\)2n

\(\Leftrightarrow\)2(n + 5) \(⋮\)2n

\(\Leftrightarrow\)2n + 10 \(⋮\)2n

\(\Leftrightarrow\)10 \(⋮\)2n

\(\Leftrightarrow\)2n \(\in\)Ư(10) = {-1; 1; -2; 2; -5; 5; -10; 10}

\(\Leftrightarrow\)n \(\in\){1; 5}

b) 2n + 7 \(⋮\)3n + 1

\(\Leftrightarrow\)3( 2n + 7)\(⋮\)3n + 1

\(\Leftrightarrow\)6n + 21\(⋮\)3n + 1

\(\Leftrightarrow\)2(3n + 1) + 19 \(⋮\)3n + 1

\(\Leftrightarrow\)19 \(⋮\)3n +1

\(\Leftrightarrow\)3n + 1 \(\in\)Ư(19) = {-1; 1; -19; 19}

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