a) 5 chia hết ( n+2)

=) ( n+2) thuộc Ư(5); n+2 thuộc ( 1; 5)

n+2=1

n= 1-2

n= -1

n+2= 5

n= 5-2

n=3

Vậy n thuộc ( -1; 3)

\(a,\Rightarrow n-2+5⋮n-2\\ \Rightarrow n-2\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Rightarrow n\in\left\{-3;1;3;7\right\}\\ b,\Rightarrow2\left(n-4\right)+13⋮n-4\\ \Rightarrow n-4\inƯ\left(13\right)=\left\{-13;-1;1;13\right\}\\ \Rightarrow n\in\left\{-9;3;5;17\right\}\\ c,\Rightarrow6n-9⋮3n+1\\ \Rightarrow2\left(3n+1\right)-12⋮3n+1\\ \Rightarrow3n+1\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\\ \Rightarrow n\in\left\{-1;0;1\right\}\left(n\in Z\right)\\ d,\Rightarrow n^2+2n-n-2+3⋮n+2\\ \Rightarrow n\left(n+2\right)-\left(n+2\right)+3⋮n+2\\ \Rightarrow n+2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Rightarrow n\in\left\{-5;-3;-1;1\right\}\)

a: \(A=\left(1-\frac49\right)\left(1-\frac{4}{25}\right)\left(1-\frac{4}{49}\right)\cdot\ldots\cdot\left(1-\frac{4}{\left(2n+1\right)^2}\right)\)

\(=\left(1-\frac23\right)\cdot\left(1-\frac25\right)\cdot\ldots\cdot\left(1-\frac{2}{2n+1}\right)\left(1+\frac23\right)\left(1+\frac25\right)\cdot\ldots\cdot\left(1+\frac{2}{2n+1}\right)\)

\(=\frac13\cdot\frac35\cdot\ldots\cdot\frac{2n-1}{2n+1}\cdot\frac53\cdot\frac75\cdot\ldots\cdot\frac{2n+3}{2n+1}\)

\(=\frac{1}{2n+1}\cdot\frac{2n+3}{3}=\frac{2n+3}{3\left(2n+1\right)}\)

b: Ta có công thức tổng quát:

\(1+\frac{1}{n^2-1}\)

\(=\frac{n^2-1+1}{n^2-1}=\frac{n^2}{n^2-1}=\frac{n\cdot n}{\left(n-1\right)\left(n+1\right)}\)

\(B=\left(1+\frac13\right)\left(1+\frac18\right)\cdot\ldots\left(1+\frac{1}{n^2-1}\right)\)

\(=\left(1+\frac{1}{2^2-1}\right)\left(1+\frac{1}{3^2-1}\right)\cdot\ldots\cdot\left(1+\frac{1}{n^2-1}\right)\)

\(=\frac{2\cdot2}{\left(2-1\right)\left(2+1\right)}\cdot\frac{3\cdot3}{\left(3-1\right)\left(3+1\right)}\cdot\ldots\cdot\frac{n\cdot n}{\left(n-1\right)\left(n+1\right)}\)

\(=\frac{2\cdot3\cdot\ldots\cdot n}{1\cdot2\cdot\ldots\cdot\left(n-1\right)}\cdot\frac{2\cdot3\cdot\ldots\cdot n}{3\cdot4\cdot\ldots\cdot\left(n+1\right)}=\frac{n}{1}\cdot\frac{2}{n+1}=\frac{2n}{n+1}\)

c: Ta có công thức tổng quát:

\(1-\frac{1}{1+2+\cdots+n}\)

\(=1-\frac{1}{\frac{n\left(n+1\right)}{2}}\)

\(=1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)

\(C=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\cdot\ldots\cdot\left(1-\frac{1}{1+2+\cdots+n}\right)\)

\(=\frac{\left(2+2\right)\left(2-1\right)}{2\left(2+1\right)}\cdot\frac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}\cdot\ldots\cdot\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)

\(=\frac{4\cdot5\cdot\ldots\cdot\left(n+2\right)}{3\cdot4\cdot\ldots\cdot\left(n+1\right)}\cdot\frac{1\cdot2\cdot\ldots\cdot\left(n-1\right)}{2\cdot3\cdot\ldots\cdot n}=\frac{n+2}{3}\cdot\frac{1}{n}=\frac{n+2}{3n}\)

giúp mik với

đáp án là B

Giá trị của số tự nhiên n trong hằng đẳng thức 𝑎^𝑛 − 𝑏^𝑛 bằng:

A. 0.

B. 1.

C. 2.

D. 3.

a: \(\Leftrightarrow n-2\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{3;1;7;-3\right\}\)

\(n+1;n;n-1\)

n+1;n;n−1