a) \(25⋮n+2\left(n\in Z\right)\)

\(\Rightarrow n+2\in\left\{-1;1;-5;5;-25;25\right\}\)

\(\Rightarrow n\in\left\{-3;-1;-7;3;-27;23\right\}\)

b) \(2n+4⋮n-1\)

\(\Rightarrow2n+4-2\left(n-1\right)⋮n-1\)

\(\Rightarrow2n+4-2n+2⋮n-1\)

\(\Rightarrow6⋮n-1\)

\(\Rightarrow n-1\in\left\{-1;1;-2;2;-3;3;-6;6\right\}\)

\(\Rightarrow n\in\left\{0;2;-1;3;-2;4;-5;7\right\}\)

c) \(1-4n⋮n+3\)

\(\Rightarrow1-4n+4\left(n+3\right)⋮n+3\)

\(\Rightarrow1-4n+4n+12⋮n+3\)

\(\Rightarrow13⋮n+3\)

\(\Rightarrow n+3\in\left\{-1;1;-13;13\right\}\)

\(\Rightarrow n\in\left\{-4;-2;-15;10\right\}\)

a) n ϵ{−3;−1;−7;3;−27;23}

b) n ∈{0;2;−1;3;−2;4;−5;7}

c) n ϵ {−4;−2;−15;10}

1) Số số hạng là n 

Tổng bằng : \(\frac{n\left(n+1\right)}{2}=378\\ \Rightarrow n\left(n+1\right)=756\\ \Rightarrow n\left(n+1\right)=27.28\\ \Rightarrow n=27\)

2) a) \(n+2⋮n-1\\ \Rightarrow n-1+3⋮n-1\\ \Rightarrow3⋮n-1\)

b) \(2n+7⋮n+1\\ \Rightarrow2\left(n+1\right)+5⋮n+1\\ \Rightarrow5⋮n+1\)

c) \(2n+1⋮6-n\\ \Rightarrow2\left(6-n\right)+13⋮6-n\\ \Rightarrow13⋮6-n\)

d) \(4n+3⋮2n+6\\ \Rightarrow2\left(2n+6\right)-9⋮2n+6\\ \Rightarrow9⋮2n+6\)

Ta có: 2n²+5n-1

=(2n²+2n+2n)+n-1

=2n​(n+2)+n-1

=(2n-1)(2n+2)

Vì 2n-1chia hết cho 2n-1 nên suy ra (2n-1)(2n+2) chia hết cho 2n-1​

Vậy 2n²+5n-1 chia hết cho 2n-1

Ta có:

\(2n^2-n+2\)

\(=2n^2+n-2n-1+3\)

\(=n.\left(2n+1\right)-\left(2n+1\right)+3\)

\(\Rightarrow n.\left(2n+1\right)⋮\left(2n+1\right)\)

\(\Rightarrow2n+1⋮2n+1\)

\(\Rightarrow3⋮2n+1\)

\(\Rightarrow2n+1\inƯC\left(3\right).\)

\(\Rightarrow2n+1\in\left\{1;-1;3;-3\right\}.\)

Có 4 trường hợp:

\(\Rightarrow\left[{}\begin{matrix}2n+1=1\\2n+1=-1\\2n+1=3\\2n+1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2n=0\\2n=-2\\2n=2\\2n=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}n=0\\n=-1\\n=1\\n=-2\end{matrix}\right.\)

Vậy \(n\in\left\{0;-1;1;-2\right\}.\)

Chúc bạn học tốt!

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