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}

c. n + 8 \(⋮\) n + 3

n + 3 + 5 \(⋮\) n + 3

\(\Rightarrow\left\{{}\begin{matrix}n+3\text{​​}⋮n+3\\5⋮n+3\end{matrix}\right.\)

\(\Rightarrow\) 5 \(⋮\) n + 3

\(\Rightarrow\) n + 3 \(\in\) Ư (5) = {1; 5}

\(\Rightarrow\) n = 2

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}

\(\Rightarrow\) 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

hay 

c: ⇔n+3+5⋮n+3⇔n+3+5⋮n+3

hay 

d: ⇔2n+2+1⋮n+1

100 thi phai

ư(29)={ 1; 29 }

n-3=1   => 4

n-3=29  => 32

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 \(n\in N\Rightarrow\left(n+4\right)\inƯ\left(11\right)=\left\{11\right\}\)

\(\Rightarrow n\in\left\{7\right\}\)

 \(19:\left(n+2\right)\)

⇒ (n+2)∈Ư(19)=(1,19)

n+2            1               19

n               -1(L)           17(TM)

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 16:

a: \(\left(x+5\right)\left(y-3\right)=15\)

=>\(\left(x+5\right)\left(y-3\right)=1\cdot15=15\cdot1=\left(-1\right)\cdot\left(-15\right)=\left(-15\right)\cdot\left(-1\right)=3\cdot5=5\cdot3=\left(-3\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-3\right)\)

=>\(\left(x+5;y-3\right)\in\left\{\left(1;15\right);\left(15;1\right);\left(-1;-15\right);\left(-15;-1\right);\left(3;5\right);\left(5;3\right);\left(-3;-5\right);\left(-5;-3\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-4;18\right);\left(10;4\right);\left(-6;-12\right);\left(-20;2\right);\left(-2;8\right);\left(0;6\right);\left(-8;-2\right);\left(-10;0\right)\right\}\)

mà (x,y) là cặp số tự nhiên

nên \(\left(x,y\right)\in\left\{\left(10;4\right);\left(0;6\right)\right\}\)

b: x là số tự nhiên

=>2x-1 lẻ và 2x-1>=-1

\(\left(2x-1\right)\left(y+2\right)=24\)

mà 2x-1>=-1 và 2x-1 lẻ

nên \(\left(2x-1\right)\cdot\left(y+2\right)=\left(-1\right)\cdot\left(-24\right)=1\cdot24=3\cdot8\)

=>\(\left(2x-1;y+2\right)\in\left\{\left(-1;-24\right);\left(1;24\right);\left(3;8\right)\right\}\)

=>\(\left(2x;y\right)\in\left\{\left(0;-26\right);\left(2;22\right);\left(4;6\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(0;-26\right);\left(1;11\right);\left(2;6\right)\right\}\)

mà (x,y) là cặp số tự nhiên

nên \(\left(x,y\right)\in\left\{\left(1;11\right);\left(2;6\right)\right\}\)

c:

x,y là các số tự nhiên

=>x+3>=3 và y+2>=2

xy+2x+3y=0

=>\(xy+2x+3y+6=6\)

=>\(x\left(y+2\right)+3\left(y+2\right)=6\)

=>\(\left(x+3\right)\left(y+2\right)=6\)

mà x+3>=3 và y+2>=2

nên \(\left(x+3\right)\cdot\left(y+2\right)=3\cdot2\)

=>x=0 và y=0

d: xy+x+y=30

=>\(xy+x+y+1=31\)

=>\(x\left(y+1\right)+\left(y+1\right)=31\)

=>\(\left(x+1\right)\left(y+1\right)=31\)

\(\Leftrightarrow\left(x+1\right)\cdot\left(y+1\right)=1\cdot31=31\cdot1=\left(-1\right)\cdot\left(-31\right)=\left(-31\right)\cdot\left(-1\right)\)

=>\(\left(x+1;y+1\right)\in\left\{\left(1;31\right);\left(31;1\right);\left(-1;-31\right);\left(-31;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;30\right);\left(30;0\right);\left(-2;-32\right);\left(-32;-2\right)\right\}\)

mà (x,y) là cặp số tự nhiên

nên \(\left(x,y\right)\in\left\{\left(0;30\right);\left(30;0\right)\right\}\)

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 \(n\in\left\{1;3;9\right\}\)

\(a=\lim4^n\left(1-\left(\dfrac{3}{4}\right)^n\right)=+\infty.1=+\infty\)

\(b=\lim\left(4^n+2.2^n+1-4^n\right)=\lim2^n\left(2+\dfrac{1}{2^n}\right)=+\infty.2=+\infty\)

\(c=limn^3\left(\sqrt{\dfrac{2}{n}-\dfrac{3}{n^4}+\dfrac{11}{n^6}}-1\right)=+\infty.\left(-1\right)=-\infty\)

\(d=\lim n\left(\sqrt{2+\dfrac{1}{n^2}}-\sqrt{3-\dfrac{1}{n^2}}\right)=+\infty\left(\sqrt{2}-\sqrt{3}\right)=-\infty\)

\(e=\lim\dfrac{3n\sqrt{n}+1}{\sqrt{n^2+3n\sqrt{n}+1}+n}=\lim\dfrac{3\sqrt{n}+\dfrac{1}{n}}{\sqrt{1+\dfrac{3}{\sqrt{n}}+\dfrac{1}{n^2}}+1}=\dfrac{+\infty}{2}=+\infty\)