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: =17+100

=117

1: =>3^x=81

=>x=4

2: =>2^x=8

=>x=3

3: =>x^3=2^3

=>x=2

4: =>x^20-x=0

=>x(x^19-1)=0

=>x=0 hoặc x=1

5: =>2^x=32

=>x=5

6: =>(2x+1)^3=9^3

=>2x+1=9

=>2x=8

=>x=4

7: =>x^3=115

=>\(x=\sqrt[3]{115}\)

8: =>(2x-15)^5-(2x-15)^3=0

=>(2x-15)^3*[(2x-15)^2-1]=0

=>2x-15=0 hoặc (2x-15)^2-1=0

=>2x-15=0 hoặc 2x-15=1 hoặc 2x-15=-1

=>x=15/2 hoặc x=8 hoặc x=7

1. Tìm số tự nhiên x biết:

1) \(3^x.3=243\)

\(3^x=243:3\)

\(3^x=81\)

\(3^x=3^4\)

\(\Rightarrow x=4\)

_____

2) \(7.2^x=56\)

\(2^x=56:7\)

\(2^x=8\)

\(2^x=2^3\)

\(\Rightarrow x=3\)

_____

3) \(x^3=8\)

\(x^3=2^3\)

\(\Rightarrow x=3\)

_____

4) \(x^{20}=x\)

\(x^{20}-x=0\)

\(x\left(x^{19}-1\right)=0\)

\(\Rightarrow x=0\) hoặc \(x=1\)

5) \(2^x-15=17\)

\(2^x=17+15\)

\(2^x=32\)

\(2^x=2^5\)

\(\Rightarrow x=5\)

_____

6) \(\left(2x+1\right)^3=9.81\)

\(\left(2x+1\right)^3=729=9^3\)

\(\rightarrow2x+1=9\)

\(2x=9-1\)

\(2x=8\)

\(x=8:2\)

\(\Rightarrow x=4\)

_____

7) \(x^6:x^3=125\)

\(x^3=125\)

\(x^3=5^3\)

\(\Rightarrow x=5\)

_____

8) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=7\\x=8\end{matrix}\right.\)

_____

9) \(3^{x+2}-5.3^x=36\)

\(3^x.\left(3^2-5\right)=36\)

\(3^x.\left(9-5\right)=36\)

\(3^x.4=36\)

\(3^x=36:4\)

\(3^x=9\)

\(3^x=3^2\)

\(\Rightarrow x=2\)

_____

10) \(7.4^{x-1}+4^{x+1}=23\)

\(\rightarrow7.4^{x-1}+4^{x-1}.4^2=23\)

\(4^{x-1}.\left(7+4^2\right)=23\)

\(4^{x-1}.\left(7+16\right)=23\)

\(4^{x-1}.23=23\)

\(4^{x-1}=23:23\)

\(4^{x-1}=1\)

\(4^{x-1}=4^1\)

\(\rightarrow x-1=0\)

\(x=0+1\)

\(\Rightarrow x=1\)

Chúc bạn học tốt

TYUIUYGBGJCYYHBVJHBNGHJK.;;JHGFFDSAQWERTYUIO

a: =>3x-9+26 chia hết cho x-3

=>\(x-3\in\left\{1;-1;2;-2;13;-13;26;-26\right\}\)

=>\(x\in\left\{4;2;5;1;16;-10;29;-23\right\}\)

b: =>6x+38 chia hết cho 2x-3

=>6x-9+47 chia hết cho 2x-3

=>\(2x-3\in\left\{1;-1;47;-47\right\}\)

=>\(x\in\left\{2;1;25;-22\right\}\)

cau 1 :1,6

câu 2 : sai đề bài

cau 3 chua lam duoc 

cau 4 : chua lam duoc

cau 5 :101/10

1) 2n - 5 \(⋮\)n + 1

    2(n + 1) - 7 \(⋮\)n + 1

Do 2(n+1) \(⋮\)n+1 nên 7 \(⋮\)n+1 \(\Rightarrow\)n + 1 \(\in\)Ư(7) = { 1; -1; 7; -7}

Với n + 1 = 1 \(\Rightarrow\)n = 0

     n + 1 = -1 \(\Rightarrow\)n = -2

     n + 1 = 7 \(\Rightarrow\)n = 6

     n + 1 = -7 \(\Rightarrow\)n = -8

Vậy n = { 0; -2; 6; -8}

a) \(\left(x+3\right)\left(2x-1\right)-\left(x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow2x^2+5x-3-x^2+2x+3=0\)

\(\Leftrightarrow x^2+7x=0\Leftrightarrow x\left(x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)

b) \(\left(x+4\right)\left(2x-3\right)-3\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow2x^2+5x-12-3x^2+12=0\)

\(\Leftrightarrow x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)