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a. n + 4 \(⋮\) n
\(\Rightarrow\left\{{}\begin{matrix}n⋮n\\4⋮n\end{matrix}\right.\)
4 \(⋮\) n
\(\Rightarrow\) n \(\in\) Ư (4) = {1; 2; 4}
\(\Rightarrow\) n \(\in\) {1; 2; 4}
b. 3n + 11 \(⋮\) n + 2
3n + 6 + 5 \(⋮\) n + 2
3(n + 2) + 5 \(⋮\) n + 2
\(\Rightarrow\left\{{}\begin{matrix}3\left(n+2\right)\text{}⋮n+2\\5⋮n+2\end{matrix}\right.\)
\(\Rightarrow\) 5 \(⋮\) n + 2
\(\Rightarrow\) n + 2 \(\in\) Ư (5) = {1; 5}
| n + 2 | 1 | 5 |
| n | vô lí | 3 |
\(\Rightarrow\) n = 3
b: \(\Leftrightarrow3n+6+5⋮n+2\)
\(\Leftrightarrow n+2\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{-1;-3;3;-7\right\}\)
c: \(\Leftrightarrow n+3+5⋮n+3\)
\(\Leftrightarrow n+3\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{-2;-4;2;-8\right\}\)
d: \(\Leftrightarrow2n+2+1⋮n+1\)
\(\Leftrightarrow n+1\in\left\{1;-1\right\}\)
hay \(n\in\left\{0;-2\right\}\)
e: \(\Leftrightarrow n-8-4⋮n-8\)
\(\Leftrightarrow n-8\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{9;7;10;6;12;4\right\}\)
b: ⇔3n+6+5⋮n+2⇔3n+6+5⋮n+2
⇔n+2∈{1;−1;5;−5}⇔n+2∈{1;−1;5;−5}
hay n∈{−1;−3;3;−7}n∈{−1;−3;3;−7}
c: ⇔n+3+5⋮n+3⇔n+3+5⋮n+3
⇔n+3∈{1;−1;5;−5}⇔n+3∈{1;−1;5;−5}
hay n∈{−2;−4;2;−8}n∈{−2;−4;2;−8}
d: ⇔2n+2+1⋮n+1
Ta có: \(M=55+225+375+13+x\)
\(=x+668\)
a: Để \(M⋮5\) thì \(x=5k+2\)
b: Để M chia 5 dư 4 thì \(x=5k+1\)
c: Để M chia 5 dư 3 thì x=5k
a) 70 - 5(x - 3 ) = 45
5( x - 3 ) = 70 - 45 = 25
x - 3 = 25 : 5 = 5
x = 5 + 3 = 8
b) (2x - 1 )4 = 3 . 62 - 27
(2x - 1 )4 = 3 . 36 - 27
(2x - 1 )4 = 81
Ta thấy 81 = 34 vậy suy ra (2x - 1)4 = 34
Để vế trong ngoặc tròn (2x - 1 ) = 3 thì x cần bằng 2
Thử lại : 2 . 2 - 1 = 4 - 1 = 3
Vậy x = 2
c) 3x3 + 43 = 102 - 33
3x3 + 43 = 100 - 33 = 67
3x3 = 67 + 43 = 110 ( Đoạn này đề bài sai hay tao sai z :)?)
b) \(\Rightarrow\left(n+2\right)\inƯ\left(19\right)=\left\{-19;-1;1;19\right\}\)
Do \(n\in N\)
\(\Rightarrow n\in\left\{17\right\}\)
a) Do \(n\in N\)
\(\Rightarrow n\inƯ\left(15\right)=\left\{1;3;5;15\right\}\)
c) \(\Rightarrow\left(n+1\right)+8⋮\left(n+1\right)\)
Do \(n\in N\Rightarrow n\inƯ\left(8\right)=\left\{1;2;4;8\right\}\)
d) \(\Rightarrow3\left(n+1\right)+18⋮\left(n+1\right)\)
Do \(n\in N\Rightarrow\left(n+1\right)\inƯ\left(18\right)=\left\{1;2;3;6;9;18\right\}\)
\(\Rightarrow n\in\left\{0;1;2;5;8;17\right\}\)
e) \(\Rightarrow\left(n-2\right)+10⋮\left(n-2\right)\)
Do \(n\in N\Rightarrow\left(n-2\right)\inƯ\left(10\right)=\left\{-2;-1;1;2;5;10\right\}\)
\(\Rightarrow n\in\left\{0;1;3;4;7;12\right\}\)
f) \(\Rightarrow n\left(n+4\right)+11⋮\left(n+4\right)\)
Do \(n\in N\Rightarrow\left(n+4\right)\inƯ\left(11\right)=\left\{11\right\}\)
\(\Rightarrow n\in\left\{7\right\}\)
Bài 10:
a: 2x-3 là bội của x+1
=>\(2x-3⋮x+1\)
=>\(2x+2-5⋮x+1\)
=>\(-5⋮x+1\)
=>\(x+1\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{0;-2;4;-6\right\}\)
b: x-2 là ước của 3x-2
=>\(3x-2⋮x-2\)
=>\(3x-6+4⋮x-2\)
=>\(4⋮x-2\)
=>\(x-2\inƯ\left(4\right)\)
=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(x\in\left\{3;1;4;0;6;-2\right\}\)
Bài 14:
a: \(4n-5⋮2n-1\)
=>\(4n-2-3⋮2n-1\)
=>\(-3⋮2n-1\)
=>\(2n-1\inƯ\left(-3\right)\)
=>\(2n-1\in\left\{1;-1;3;-3\right\}\)
=>\(2n\in\left\{2;0;4;-2\right\}\)
=>\(n\in\left\{1;0;2;-1\right\}\)
mà n>=0
nên \(n\in\left\{1;0;2\right\}\)
b: \(n^2+3n+1⋮n+1\)
=>\(n^2+n+2n+2-1⋮n+1\)
=>\(n\left(n+1\right)+2\left(n+1\right)-1⋮n+1\)
=>\(-1⋮n+1\)
=>\(n+1\in\left\{1;-1\right\}\)
=>\(n\in\left\{0;-2\right\}\)
mà n là số tự nhiên
nên n=0
Bài 5:
b: Ta có: \(n+6⋮n+2\)
\(\Leftrightarrow n+2\in\left\{2;4\right\}\)
hay \(n\in\left\{0;2\right\}\)
c: Ta có: \(3n+1⋮n-2\)
\(\Leftrightarrow n-2\in\left\{-1;1;7\right\}\)
hay \(n\in\left\{1;3;9\right\}\)
a: Ta có: \(n+4⋮n\)
\(\Leftrightarrow4⋮n\)
\(\Leftrightarrow n\in\left\{1;-1;2;-2;4;-4\right\}\)
b: Ta có: \(3n+11⋮n+2\)
\(\Leftrightarrow n+2\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{-1;-3;3;-7\right\}\)
c: Ta có: \(n+8⋮n+3\)
\(\Leftrightarrow n+3\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{-2;-4;2;-8\right\}\)
a) \(n+4⋮n\)
Vì \(n⋮n\Rightarrow4⋮n\Rightarrow n\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
b) \(3n+11⋮n+2\\ \Rightarrow\left(3n+6\right)+5⋮n+2\\ \Rightarrow3\left(n+2\right)+5⋮n+2\)
Vì \(3\left(n+2\right)⋮n+2\Rightarrow5⋮n+2\Rightarrow n+2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\Rightarrow n\in\left\{-7;-3;-1;3\right\}\)
c) \(n+8⋮n+3\\ \Rightarrow\left(n+3\right)+5⋮n+3\)
Vì \(n+3⋮n+3\Rightarrow5⋮n+3\Rightarrow n+3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\Rightarrow n\in\left\{-8;-4;-2;2\right\}\)