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

a: \(n\in\left\{1;-1\right\}\)

a, x= 3

b, ko có x thỏa mãn

c, x= 3

d, x= 2

a: x=3

b: \(2x-1=2\)

hay \(x=\dfrac{3}{2}\)

c: \(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)

(n + 1)³ = (n + 1)²

=> (n + 1)³ - (n + 1)² = 0

=> (n + 1)².(n + 1 - 1) = 0

=> (n + 1)².n = 0

=> \(\orbr{\begin{cases}\left(n+1\right)^2=0\\n=0\end{cases}}\)=> \(\orbr{\begin{cases}n+1=0\\n=0\end{cases}}\)=> \(\orbr{\begin{cases}n=-1\\n=0\end{cases}}\)

Mà \(n\in N\Rightarrow n=0\)

a) 15 ⋮ 1 , 3 , 5, 15

Vậy n = 0 , 2 , 4 , 14

a) \(\Rightarrow\left(n+1\right)\inƯ\left(15\right)=\left\{-15;-5;-3;-1;1;3;5;15\right\}\)

Do \(n\in N\)

\(\Rightarrow n\in\left\{0;2;4;14\right\}\)

b) \(\Rightarrow\left(n+1\right)+2⋮\left(n+1\right)\)

Do \(n\in N\Rightarrow\left(n+1\right)\inƯ\left(2\right)=\left\{1;2\right\}\Rightarrow n\in\left\{0;1\right\}\)

a: \(\Leftrightarrow n+1\in\left\{1;3;5;15\right\}\)

hay \(n\in\left\{0;2;4;14\right\}\)

`a)(x+5)^3=-64`

`(x+5)^3=(-4)^3`

`x+5=-4`

`x=-4-5=-9`

Vậy `x=-9`

`2)(2x-3)^3=8`(9 không được)

`(2x-3)^3=2^3`

`2x-3=2`

`2x=5`

`x=5/2`

Vậy `x=5/2`

a)\(\left(x+5\right)^3=64\)

\(\left(x+5\right)^3=4^3\)

\(x+5=4\)

\(x=4-5\)

\(x=-1\)

b) \(\left(2x-3\right)^3=9\)

\(\left(2x-3\right)^3=3^3\)

\(2x-3=3\)

\(2x=3+3\)

\(2x=6\)

\(x=\dfrac{6}{2}\)

\(x=3\)

Giải:

a) \(\left(x+5\right)^3=-64\) 

\(\Rightarrow\left(x+5\right)^3=\left(-4\right)^3\) 

\(\Rightarrow x+5=-4\) 

\(\Rightarrow x=-4-5\) 

\(\Rightarrow x=-9\) 

b) \(\left(2x-3\right)^3=8\) 

\(\Rightarrow\left(2x-3\right)^3=2^3\) 

\(\Rightarrow2x-3=2\) 

\(\Rightarrow2x=2+3\) 

\(\Rightarrow2x=5\) 

\(\Rightarrow x=5:2\) 

\(\Rightarrow x=\dfrac{5}{2}\)

 3^2.3^n=3^5

=> n=3

(2^2:4).n^2=4

=> n= +2;-2

1/9.3^4.3^n=3^7

=>n=9

1/9=27^n=3^n

=> n=

1/2.2n+4.2^n=9.5^n

=> n ∈ ∅

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