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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) \(4n+3⋮2n+6\\ \Rightarrow2\left(2n+6\right)-9⋮2n+6\\ \Rightarrow9⋮2n+6\)
=>n^2-n+4n-4+5 chia hết cho n-1
=>\(n-1\in\left\{1;-1;5;-5\right\}\)
mà n>=0
nên \(n\in\left\{2;0;6\right\}\)
Bài 2:
a: Để E là số nguyên thì \(3n+5⋮n+7\)
\(\Leftrightarrow3n+21-16⋮n+7\)
\(\Leftrightarrow n+7\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16\right\}\)
hay \(n\in\left\{-6;-8;-5;-9;-3;-11;1;-15;9;-23\right\}\)
b: Để F là số nguyên thì \(2n+9⋮n-5\)
\(\Leftrightarrow2n-10+19⋮n-5\)
\(\Leftrightarrow n-5\in\left\{1;-1;19;-19\right\}\)
hay \(n\in\left\{6;4;29;-14\right\}\)
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}
Lời giải:
\(A=\frac{n}{n+1}+\frac{n+1}{n+2}=\frac{n(n+2)+(n+1)^2}{(n+1)(n+2)}=\frac{2n^2+4n+2}{n^2+3n+2}>1\) do $2n^2+4n+2> n^2+3n+2$ với mọi $n\in\mathbb{N}^*$
$B=\frac{2n+1}{2n+3}< 1$ do $2n+1< 2n+3$
Do đó $A>B$
a) \(n+1\inƯ\left(n^2+2n-3\right)\)
\(\Leftrightarrow n^2+2n-3⋮n+1\)
\(\Leftrightarrow n\left(n+1\right)+n-3⋮n+1\)
Vì \(n\left(n+1\right)⋮n+1\Rightarrow n-3⋮n+1\)
\(\Leftrightarrow n+1-4⋮n+1\)
Vì \(n+1⋮n+1\Rightarrow-4⋮n+1\Rightarrow n+1\inƯ\left(-4\right)=\left\{-1;1;-2;2;-4;4\right\}\)
Ta có bảng sau:
Vậy...
b) \(n^2+2\in B\left(n^2+1\right)\)
\(\Leftrightarrow n^2+2⋮n^2+1\)
\(\Leftrightarrow n^2+1+1⋮n^2+1\)
Vì \(n^2+1⋮n^2+1\) nên \(1⋮n^2+1\Rightarrow n^2+1\inƯ\left(1\right)=\left\{-1;1\right\}\)
Ta có bảng sau:
\(0\) (tm)
Vậy \(n=0\)
c) \(2n+3\in B\left(n+1\right)\)
\(\Leftrightarrow2n+3⋮n+1\)
\(\Leftrightarrow2n+2+1⋮n+1\)
\(\Leftrightarrow2\left(n+1\right)+1⋮n+1\)
Vì \(2\left(n+1\right)⋮n+1\) nên \(1⋮n+1\Rightarrow n+1\inƯ\left(1\right)=\left\{-1;1\right\}\)
Ta có bảng sau:
Vậy...
a) n+1∈Ư(n2+2n−3)n+1∈Ư(n2+2n−3)
⇔n2+2n−3⋮n+1⇔n2+2n−3⋮n+1
⇔n(n+1)+n−3⋮n+1⇔n(n+1)+n−3⋮n+1
Vì n(n+1)⋮n+1⇒n−3⋮n+1n(n+1)⋮n+1⇒n−3⋮n+1
⇔n+1−4⋮n+1⇔n+1−4⋮n+1
Vì n+1⋮n+1⇒−4⋮n+1⇒n+1∈Ư(−4)={−1;1;−2;2;−4;4}n+1⋮n+1⇒−4⋮n+1⇒n+1∈Ư(−4)={−1;1;−2;2;−4;4}
Ta có bảng sau:
Vậy...
b) n2+2∈B(n2+1)n2+2∈B(n2+1)
⇔n2+2⋮n2+1⇔n2+2⋮n2+1
⇔n2+1+1⋮n2+1⇔n2+1+1⋮n2+1
Vì n2+1⋮n2+1n2+1⋮n2+1 nên 1⋮n2+1⇒n2+1∈Ư(1)={−1;1}1⋮n2+1⇒n2+1∈Ư(1)={−1;1}
Ta có bảng sau:
00 (tm)
Vậy n=0n=0
c) 2n+3∈B(n+1)2n+3∈B(n+1)
⇔2n+3⋮n+1⇔2n+3⋮n+1
⇔2n+2+1⋮n+1⇔2n+2+1⋮n+1
⇔2(n+1)+1⋮n+1⇔2(n+1)+1⋮n+1
Vì 2(n+1)⋮n+12(n+1)⋮n+1 nên 1⋮n+1⇒n+1∈Ư(1)={−1;1}1⋮n+1⇒n+1∈Ư(1)={−1;1}
Ta có bảng sau: