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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) \(x+1^3=2^5-\left(-1^3\right)\)

\(\Rightarrow x+1=33\)

=> x = 32

b) \(3^7-x=1^4-\left(-3^5\right)\)

\(\Rightarrow2187-x=1+243=244\)

=> x = 1943

a) \(\Leftrightarrow x+1=32+1\)

\(\Leftrightarrow x=32\)

Vậy x = 32

b) \(\Leftrightarrow2187-x=1+243\)

\(\Leftrightarrow2187-x=244\)

\(\Leftrightarrow x=1943\)

Vậy x = 1943

a) \(x+1^3=2^5-\left(-1^3\right)\) 

      \(x+1=32-\left(-1\right)\) 

      \(x+1=33\) 

            \(x=33-1\) 

            \(x=32\) 

b) \(3^7-x=1^4-\left(-3^5\right)\) 

\(2187-x=1-\left(-243\right)\) 

\(2187-x=244\) 

            \(x=2187-244\) 

            \(x=1943\)

b. (x + 4)² - (x + 1)(x - 1) = 16

<=> x² + 4x + 16 - (x² - 1) = 16

<=> x² + 4x + 16 - x² + 1 - 16 = 0

<=> x² - x² + 4x = 16 - 16 - 1

<=> 4x = -1

<=> x = \(\dfrac{-1}{4}\)

\(a,\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\\ \Leftrightarrow48x=46\\ \Leftrightarrow x=\dfrac{23}{24}\\ b,\Leftrightarrow x^2+8x+16-x^2+1=16\\ \Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)

a: \(x\in\left\{1;7\right\}\)

b: \(x+1=1\)

hay x=0

\(a,\Rightarrow x\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ b,\Rightarrow2\left(x+1\right)-1⋮x+1\\ \Rightarrow x+1\inƯ\left(1\right)=\left\{-1;1\right\}\\ \Rightarrow x\in\left\{-2;0\right\}\)

a: 

b: 

hay x=0

Bài 4:

a) \(\dfrac{x}{2}=\dfrac{2}{4}\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{1}{2}\)

\(\Rightarrow x=2\)

Vậy: \(x=2\)

b) \(-\dfrac{1}{5}=\dfrac{2}{x}\)

\(\Rightarrow x=\dfrac{-5.2}{1}=-10\)

Vậy: \(x=-10\)

c) \(\dfrac{x}{5}=\dfrac{5}{x}\)

\(\Rightarrow x^2=25\)

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

Vậy: \(x\in\left\{5;-5\right\}\)

\(\dfrac{x}{2}=\dfrac{2}{4}\\ =>x=\dfrac{2.2}{4}=1\)

\(\dfrac{-1}{5}=\dfrac{2}{x}\\ =>x=\dfrac{2.5}{-1}=-10\)

\(\dfrac{x}{5}=\dfrac{5}{x}\\ =>x^2=25\\ x=5;x=-5\)

a) 2

b) 5,-5

a)\(3x\left(x-1\right)+2x^2\left(x-1\right)=0\\ \Leftrightarrow x\left(x-1\right)\left(3+2x\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\x-1=0\\3+2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\\x=\dfrac{-3}{2}\end{matrix}\right.\)

a: Ta có: \(3x^2-3x+2x^3-2x^2=0\)

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

\(\Leftrightarrow x\left(2x^2+x-3\right)=0\)

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

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

b: Ta có: \(x^3+27=-x^2+9\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-3\right)=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

Bài 1 :

Gọi \(A=5+5^2+5^3+...+5^{98}+5^{99}\\ 5A=5^2+5^3+5^4+...+5^{99}+5^{100}\\ 5A-A=\left(5^2+5^3+5^4+...+5^{99}+5^{100}\right)-\left(5+5^2+5^3+...+5^{98}+5^{99}\right)\\ 4A=5^{100}-5\\ A=\dfrac{5^{100}-5}{4}\)

Bài 2:

\(\left(12x-4\right)\cdot8^{2022}=4\cdot8^{2023}\\ 12x-4=4\cdot8^{2023}:8^{2022}\\ 12x-4=4\cdot8\\ 12x-4=32\\ 12x=36\\ x=3\)