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\(1) VP= \frac{1}{n}-\frac{1}{n+1}\)\(= \frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}\)\(= \frac{n+1-n}{n(n+1)}\)\(= \frac{1}{n(n+1)}\)\(= VT\)
2) \(VP= \frac{1}{n+1}-\frac{1}{(n+1)(n+2)}= \frac{(n+2)}{n(n+1)(n+2)}-\frac{n}{n(n+1)(n+2)}\)\(= \frac{n+2-n}{n(n+1)(n+2)}= \frac{2}{n(n+1)(n+2)}=VT\)
3) \(VP= \frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}=\frac{n+3}{n(n+1)(n+2)(n+3)}-\frac{n}{n(n+1)(n+2)(n+3)}\)\(= \frac{n+3-n}{n(n+1)(n+2)(n+3)}=\frac{3}{n(n+1)(n+2)(n+3)(n+4)}=VT\)
Những ý sau làm tương tự, thế mà chẳng thèm mở mồm ra hỏi bạn :))
a) Ta có : n+1⋮ n+1
⇒[(n+6)-(n+1)]⋮n+1
⇒5⋮n+1
⇒n+1ϵ {-1;1;5;-5}
⇒nϵ{0;-2;4;-6}
b) Ta có :2(2n+1)⋮2n+1⇔4n+2⋮2n+1
Mà 4n+9⋮2n+1
⇒[(4n+9)-(4n+2)]⋮2n+1
⇒7⋮2n+1⇔2n+1ϵ{-1;1;-7;7}
|
2n+1 |
1 | -1 | -7 | 7 |
| 2n | 0 | -2 | -8 | 6 |
| n | 0 | -1 | -4 | 3 |
c)Ta có : 2(n-1)⋮n-1⇔2n-2⋮n-1
⇒[(2n)-(2n-2)]⋮n-1
⇒2⋮n-1⇔n-1ϵ{1;-1;-2;2}
| n-1 | 1 | -1 | 2 | -2 |
| n | 2 | 0 | 3 | -1 |
d)n+4⋮n+1
⇒[(n+4)-(n+1)]⋮n+1
⇒3⋮n+1⇔n+1ϵ{1;-1;3;-3}
| n+1 | 1 | -1 | 3 | -3 |
| n | 0 | -2 | 2 | -4 |
a) Ta chia làm 2 trường hợp
*Trường hơp 1: n chẵn
Nếu n chẵn => (n + 10)⋮2 => (n+10)(n+15)⋮2
*Trường hợp 2: n lẻ
Nếu n lẻ => (n + 15)⋮ 2 => (n+10)(n+15)⋮2
Vậy với mọi trường hợp n ∈ N thì (n+10)(n+15)⋮2
6/ \(\frac{2n-4}{n}=\frac{2n}{n}-\frac{4}{n}\) \(=2-\frac{4}{n}\)
Để 2n - 4 chia hết cho n thì 4 chia hết cho n
\(\Rightarrow\) n = 1; n = 2; n = 4
7/ \(\frac{35+12n}{n}=\frac{35}{n}+\frac{12n}{n}=\frac{35}{n}+12\)
Để 35 + 12n chia hết cho n thì 35 chia hết cho n
\(\Rightarrow\) n = 1; n = 5; n = 7; n = 35
1/ Để 7 \(⋮\) n (n \(\in N\)) thì n = 1; n = 7
2/ Để 7 \(⋮\) \(\left(n-1\right)\) thì \(n-1=1;n-1=-1;n-1=7;n-1=-7\)
*) \(n-1=1\)
n = 1 + 1
n = 2 (thỏa mãn n là số tự nhiên)
*) \(n-1=-1\)
\(n=-1+1\)
n = 0 (thỏa mãn n là số tự nhiên)
*) \(n-1=7\)
n = 7 + 1
n = 8 (thỏa mãn n là số tự nhiên)
*) \(n-1=-7\)
\(n=-7+1\)
\(n=-6\) (không thỏa mãn n là số tự nhiên)
Vậy n = 8; n = 2; n = 0
a, Ta có : \(\text{n + 5 = (n - 1)+6}\)
Vì \(\text{(n-1) ⋮ n-1}\)
Nên để \(\text{n+5 ⋮ n-1}\)⋮ `n-1`
Thì \(\text{6 ⋮ n-1}\)
\(\Rightarrow\) \(\text{n - 1 ∈ Ư(6)}\)
\(\Rightarrow\) \(\text{n - 1 ∈}\) \(\left\{\text{±1;±2;±3;±6}\right\}\)
\(\Rightarrow\) \(\text{n ∈}\) \(\left\{\text{0;-1;-2;-5;2;3;4;7}\right\}\) \(\text{( TM )}\)
\(\text{________________________________________________________}\)
b, Ta có : \(\text{2n-4 = (2n+4)- 8 = 2(n+2) - 8}\)
Vì \(\text{2(n+2) ⋮ n+2}\)
Nên để \(\text{2n-4 ⋮ n+2}\)
Thì \(\text{8 ⋮ n+2}\)
\(\Rightarrow\) \(\text{n + 2 ∈ Ư(8)}\)
\(\Rightarrow\) \(\text{n + 2 ∈}\) \(\left\{\text{±1;±2;±4;±8}\right\}\)
\(\Rightarrow\) \(\text{n ∈}\) \(\left\{\text{-3;-4;-6;-10;-1;0;2;6}\right\}\) ( TM )
\(\text{_________________________________________________________________ }\)
c, Ta có :\(\text{ 6n + 4 = (6n + 3) +1 = 3(2n+1) + 1}\)
Vì \(\text{3(2n+1) ⋮ 2n+1}\)
Nên để\(\text{ 6n+4 ⋮ 2n+1}\)
Thì \(\text{1 ⋮ 2n+1}\)
\(\Rightarrow\) \(\text{2n + 1 ∈ Ư(1)}\)
\(\Rightarrow\) \(\text{2n + 1 ∈}\) \(\left\{\text{±1}\right\}\)
\(\Rightarrow\) \(\text{2n ∈}\) \(\left\{\text{-2;0}\right\}\)
\(\Rightarrow\) \(\text{n ∈}\) \(\left\{\text{-1;0}\right\}\) ( TM )
\(\text{_______________________________________}\)
Ta có : \(\text{3 - 2n = -( 2n - 3 ) = -( 2n + 2 ) + 5 = -2( n+1)+5}\)
Vì \(\text{-2(n+1) ⋮ n+1}\)
Nên để \(\text{3-2n ⋮ n+1}\)
Thì\(\text{ 5 ⋮ n + 1}\)
\(\Rightarrow\) \(\text{n + 1 ∈}\) \(\left\{\text{±1;±5}\right\}\)
\(\Rightarrow\) \(\text{n ∈}\) \(\text{-2;-6;0;4}\) ( TM )
\(1.2+2.3+3.4+...+n\left(n+1\right)=\frac{1.2.3+2.3.3+3.4.3+...+n\left(n+1\right).3}{3}\)
\(=\frac{1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]}{3}\)
\(=\frac{1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)}{3}\)
\(=\frac{n\left(n+1\right)\left(n+2\right)}{3}=\frac{n\left(n+1\right)\left(2n+4\right)}{6}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{3n\left(n+1\right)}{6}\)
\(=\frac{n\left(n+1\right)\left(2n+1\right)}{6}+\frac{n\left(n+1\right)}{2}\)
Vậy chọn C
c c c c c cccccccc c c c cccc cccccc ccccccccc ccccccccccccccccccc cc