a: =>1/3x-2/5x=5

=>-1/15x=5

=>x=-75
b: =>4x=4

=>x=1

c: =>6*3^x-5*3^x=243

=>3^x=243

=>x=5

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^3=343=7^3`

`=>x=7`

Vậy `x=7`

`b)(x-2,5)^4=(x-2,5)^2`

`=>(x-2,5)^2[(x-2,5)^2-1]=0`

`+)(x-2,5)^2=0<=>x=2,5`

`+)(x-2,5)^2=1`

`TH1:x-2,5=1<=>x=3,5`

`th2:x-2,5=-1<=>x=1,5`

Vậy `x=0` hoặc `x=1,5` hoặc `x=3,5

`c)x^8/243=27`

`=>x^8=27.243`

`=>x^8=3^3*3^5=3^8`

`=>x=+-3`

Giải:

a) \(x^3=343\) 

\(\Rightarrow x^3=7^3\) 

\(\Rightarrow x=7\) 

b) \(\left(x-2,5\right)^4=\left(x-2,5\right)^2\) 

Vì có cùng cơ số nhưng lại khác mũ thì \(\left(x-2,5\right)\in\left\{0;1\right\}\) 

TH1:

\(\Rightarrow x-2,5=0\) 

\(\Rightarrow x=2,5\) 

TH2:

\(\Rightarrow\left[{}\begin{matrix}x-2,5=1\\x-2,5=-1\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=3,5\\x=1,5\end{matrix}\right.\) 

c) \(\dfrac{x^8}{243}=27\) 

\(\Rightarrow x^8=3^3.3^5\) 

\(\Rightarrow x^8=3^8\) 

\(\Rightarrow x=\pm3\)

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 \(n+3⋮n+3\Rightarrow5⋮n+3\Rightarrow n+3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\Rightarrow n\in\left\{-8;-4;-2;2\right\}\)

M=\(-2x^2y+xy^2-4x+4xy-2xy^2+3x^2y-4xy\)

\(M=x^2y-xy^2-4x\)

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

a) \(\dfrac{1}{8}.16^n=2^n\)

=>\(\dfrac{2^{4n}}{2^3}=2^n\)

=>\(2^{4n-3}=2^n\)

=>4n-3=n

=>3n-3=0

=>n=1.

b) \(27< 3^n< 243\)

=>\(3^3< 3^n< 3^5\). Mà n là số tự nhiên.

- Vậy n=4

a) \(\dfrac{1}{8}.16^n=2^n\)

\(\Rightarrow2^{4n}=2^3.2^n\)

\(\Rightarrow4n=3+n\)

\(\Rightarrow3n=3\)

\(\Rightarrow n=1\)

Vậy: \(n=1\)

b) \(27< 3^n< 243\)

\(\Rightarrow3^3< 3^n< 3^5\)

\(\Rightarrow3< n< 5\)

\(\Rightarrow n=4\)

Vậy: \(n=4\)

(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) 5^3 = 125

b)3^4.3^3=3^7

c)27.3^2=243

d)49.7^2=2401

e) chịu -_-