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a)\(n+7⋮n+2\)
\(\Rightarrow\left(n+2\right)+5⋮n+2\)
\(\Rightarrow5⋮n+2\)
\(\Rightarrow n+2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta có bảng sau:
| n+2 | 1 | -1 | 5 | -5 |
| n | -1 | -3 | 3 | -7 |
Vậy \(n\in\left\{-1;-3;3;-7\right\}\)
b)\(9-n⋮n-3\)
\(\Rightarrow6-\left(n-3\right)\)
\(\Rightarrow6⋮n-3\)
\(\Rightarrow n-3\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
nếu n-3=1 thì n=4
nếu n-3=-1 thì n=2
nếu n-3=2 thì n=5
nếu n-3=-2 thì n=1
nếu n-3=3 thì n=6
nếu n-3=-3 thì n=0
nếu n-3=6 thì n=9
nếu n-3=-6 thì n=-3
Vậy \(n\in\left\{4;2;5;1;6;0;9;-3\right\}\)
c)\(n^2+n+17⋮n+1\)
\(\Rightarrow n\left(n+1\right)+17⋮n+1\)
\(\Rightarrow17⋮n+1\)
\(\Rightarrow n+1\inƯ\left(17\right)=\left\{\pm1;\pm17\right\}\)
nếu n+1=1 thì n=0
nếu n+1=-1 thì n=-2
nếu n+1=17 thì n=16
nếu n+1=-17 thì n=-18
Vậy \(n\in\left\{0;-2;16;-18\right\}\)
\(a,n+9⋮n+2\)
\(\Rightarrow n+2+7⋮n+2\)
mà \(n+2⋮n+2\Rightarrow n+2\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(n\in\left\{-1;-3;5;-9\right\}\)
\(b,2n+7⋮n+1\)
\(\Rightarrow2n+2+5⋮n+1\)
\(\Rightarrow2\left(n+1\right)+5⋮n+1\)
mà \(2\left(n+1\right)⋮n+1\Rightarrow5⋮n+1\)
\(\Rightarrow n+1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Rightarrow n\in\left\{0;-2;4;-6\right\}\)
a, n + 2 \(⋮n-3\)
<=> n - 3 + 5 \(⋮n-3\)
<=> 5 \(⋮n-3\)
=> n - 3 \(\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
=> n = 4; 2; 8; -2 (thỏa mãn)
b, 3n + 15 \(⋮n-4\)
Có 3(n - 4) \(⋮n-4\)
=> (3n + 15) - (3n - 12) \(⋮n-4\)
<=> 27 \(⋮n-4\)
=> n - 4 \(\inƯ\left(27\right)=\left\{\pm1;\pm3;\pm9;\pm27\right\}\)
=> n = 5; 3; 7; 1; 13; -5; 31; -23 (thỏa mãn)
@hoang thuy an
c, 2n - 3 \(⋮3n+2\)
<=> 3(2n - 3) \(⋮3n+2\)
<=> 6n - 9 \(⋮3n+2\)
Có 2(3n + 2) \(⋮3n+2\)
=> (6n - 9) - (6n + 4) \(⋮3n+2\)
<=> -13 \(⋮3n+2\)
=> 3n + 2 \(\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
=> 3n = -1; -3; 11; -15
=> n = -\(\dfrac{1}{3};-1;\dfrac{11}{3};-5\)
Mà n \(\in Z\Rightarrow n=-1;-5\)
d, 4n + 7 \(⋮3n+1\)
<=> 3(4n + 7) \(⋮3n+1\)
<=> 12n + 21 \(⋮3n+1\)
Có 4(3n + 1) \(⋮3n+1\)
=> (12n + 21) - (12n + 4) \(⋮3n+1\)
<=> 17 \(⋮3n+1\)
=> 3n + 1 \(\inƯ\left(17\right)=\left\{\pm1;\pm17\right\}\)
=> 3n = 0; -2; 16; -18
=> n = 0; -\(\dfrac{2}{3};\dfrac{16}{3};-6\)
Mà n \(\in Z\Rightarrow n=0;-6\)
@hoang thuy an
a) => n-1+3 chia hết n-1
Mà n-1 chia hết n-1
=> 3 chia hết cho n-1
=> n-1 thuộc Ước của 3
........
b)=> 2(n+1) +5 chia hết n+1
mà 2(n+1) chia hết n+1
=> 5 chia hết cho n+1
=> n+1 thuộc ước của 5
.......
a,Ta có :\(n+2⋮n-1\)
\(=>n-1+3⋮n-1\)
Do \(n-1⋮n-1\)
\(=>3⋮n-1\)
\(=>n-1\inƯ\left(3\right)\)
\(=>n-1\in\left\{-3;-1;1;3\right\}\)
\(=>n\in\left\{-2;0;2;4\right\}\)
b,\(2n+7⋮n+1\)
\(=>2.\left(n+1\right)+5⋮n+1\)
Do \(2.\left(n+1\right)⋮n+1\)
\(=>5⋮n+1\)
\(=>n+1\inƯ\left(5\right)\)
\(=>n+1\in\left\{-5;-1;1;5\right\}\)
\(=>n\in\left\{-6;-2;0;4\right\}\)
Bài 1:
b) Ta có:
\(16^5=2^{20}\)
\(\Rightarrow B=16^5+2^{15}=2^{20}+2^{15}\)
\(\Rightarrow B=2^{15}.2^5+2^{15}\)
\(\Rightarrow B=2^{15}\left(2^5+1\right)\)
\(\Rightarrow B=2^{15}.33\)
\(\Rightarrow B⋮33\) (Đpcm)
c) \(C=5+5^2+5^3+5^4+...+5^{100}\)
\(\Rightarrow C=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{99}+5^{100}\right)\)
\(\Rightarrow C=1\left(5+5^2\right)+5^2\left(5+5^2\right)+...+5^{98}\left(5+5^2\right)\)
\(\Rightarrow\left(1+5^2+...+5^{98}\right)\left(5+5^2\right)\)
\(\Rightarrow C=Q.30\)
\(\Rightarrow C⋮30\) (Đpcm)
Bài 1 : a, \(A=1+3+3^2+...+3^{118}+3^{119}\)
\(A=\left(1+3+3^2+3^3\right)+...+\left(3^{116}+3^{117}+3^{118}+3^{119}\right)\)
\(A=\left(1+3+3^2+3^3\right)+...+3^{116}\left(1+3+3^2+3^3\right)\)
\(A=1.30+...+3^{116}.30=\left(1+...+3^{116}\right).30⋮3\)
Vậy \(A⋮3\)
b, \(B=16^5+2^{15}=\left(2.8\right)^5+2^{15}\)
\(=2^5.8^5+2^{15}=2^5.\left(2^3\right)^5+2^{15}\)
\(=2^5.2^{15}+2^{15}.1=2^{15}\left(32+1\right)=2^{15}.33⋮33\)
Vậy \(B⋮33\)
c, Tương tự câu a nhưng nhóm 2 số
Bài 2 : a, \(n+2⋮n-1\) ; Mà : \(n-1⋮n-1\)
\(\Rightarrow\left(n+2\right)-\left(n-1\right)⋮n-1\)
\(\Rightarrow n+2-n+1⋮n-1\Rightarrow3⋮n-1\)
\(\Rightarrow n-1\in\left\{1;3\right\}\Rightarrow n\in\left\{2;4\right\}\)
Vậy \(n\in\left\{2;4\right\}\) thỏa mãn đề bài
b, \(2n+7⋮n+1\)
Mà : \(n+1⋮n+1\Rightarrow2\left(n+1\right)⋮n+1\Rightarrow2n+2⋮n+1\)
\(\Rightarrow\left(2n+7\right)-\left(2n+2\right)⋮n+1\)
\(\Rightarrow2n+7-2n-2⋮n+1\Rightarrow5⋮n+1\)
\(\Rightarrow n+1\in\left\{1;5\right\}\Rightarrow n\in\left\{0;4\right\}\)
Vậy \(n\in\left\{0;4\right\}\) thỏa mãn đề bài
c, tương tự phần b
d, Vì : \(4n+3⋮2n+6\)
Mà : \(2n+6⋮2n+6\Rightarrow2\left(2n+6\right)⋮2n+6\Rightarrow4n+12⋮2n+6\)
\(\Rightarrow\left(4n+12\right)-\left(4n+3\right)⋮2n+6\)
\(\Rightarrow4n+12-4n-3⋮2n+6\Rightarrow9⋮2n+6\)
\(\Rightarrow2n+6\in\left\{1;2;9\right\}\Rightarrow2n=3\Rightarrow n\in\varnothing\)
Vậy \(n\in\varnothing\)
\(a,\frac{n+5}{n+2}=\frac{n+2+3}{n+2}=1+\frac{3}{n+2}\)
Để \(n+5⋮n+2\) thì \(n+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Xét bảng ( tự xét nha )
KL..
\(b,\frac{2n+3}{n-2}=\frac{2\left(n-2\right)+7}{n-2}=2+\frac{7}{n-2}\)
Giải các ý khác tương tự như trên
Ta có n+5=n+2+3
Để n+5 chia hết cho n+2 thì n+2+3 chia hết cho n+2
Mà n thuộc n => n+2 thuộc N
=> n+2 thuộc Ư (5)={1;5}
Nếu n+2=1 => n=-1 (ktm)
Nếu n+1=5 => n=4(tm)
Vậy n=4 thì n+5 chia hết cho n+2
b) Ta có 2n+3=2(n-2)+7
Để 2n+3 chia hết cho n-2 thì 2(n-2)+7 chia hết cho n-1
n thuộc N => n-1 thuộc N
=> n-1 thuộc Ư (7)={1;7}
Nếu n-1=1 => n=2(tm)
Nếu n-1=7 => n=8 (tm)
Đề sai thì phải ! Học Lớp 7 mới giải xong bài này !
\(\frac{1}{9}\cdot27^n=3^n\)
\(\frac{1}{9}\cdot\left(3^3\right)^n=3^n\)
\(\frac{1}{9}\cdot3^{3n}=3^n\)
\(\frac{1}{9}=3^n\text{ : }3^{3n}\)
\(\frac{1}{9}=3^{-2n}\)
\(\frac{1}{3^2}=\frac{1}{3^{2n}}\)
\(\Rightarrow\text{ }3^{2n}=3^2\)
\(3^{2n}-3^2=0\)
\(3\left(3^{2n-1}-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3=0\text{ ( Vô lí ) }\\3^{2n-1}-3=0\end{cases}}\) \(\Rightarrow\text{ }3^{2n-1}=3\) \(\Rightarrow\text{ }2n-1=1\) \(\Rightarrow\text{ }2n=2\) \(\Rightarrow\text{ }n=1\)
Vậy \(n=1\)
a/ \(15⋮n-2\)
\(\Leftrightarrow n-2\inƯ\left(15\right)\)
Mà \(n\in N\)
\(\Leftrightarrow\left[{}\begin{matrix}n-2=1\\n-2=15\\n-2=5\\n-2=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n=3\\n=17\\n=7\\n=5\end{matrix}\right.\)
Vậy ............
b/ \(18⋮n+1\)
\(\Leftrightarrow n+1\inƯ\left(18\right)\)
Mà \(n\in N\)
Vậy .....
a) Ta có:
\(15⋮n-2\)
\(\Rightarrow n-2\in\left\{1;3;5;15\right\}\) ( Vì \(n\in N\))
+) \(n-2=1\Rightarrow n=3\)
+) \(n-2=3\Rightarrow n=5\)
+) \(n-2=5\Rightarrow n=7\)
+) \(n-2=15\Rightarrow n=17\)
Vậy \(n=3\) hoặc \(n=5\) hoặc \(n=7\) hoặc \(n=17\)
b) Ta có:
\(18⋮n+1\)
\(\Rightarrow n+1\in\left\{1;2;3;6;9;18\right\}\) ( Vì \(n\in N\))
+) \(n+1=1\Rightarrow n=0\)
+) \(n+1=2\Rightarrow n=1\)
+) \(n+1=3\Rightarrow n=2\)
+) \(n+1=6\Rightarrow n=5\)
+) \(n+1=9\Rightarrow n=8\)
+) \(n+1=18\Rightarrow n=17\)
Vậy \(n=0\) hoặc \(n=1\) hoặc \(n=2\) hoặc \(n=5\) hoặc \(n=8\) hoặc \(n=17\)