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)

\(\begin{array}{l}\left( {9x - {2^3}} \right):5 = 2\\9x - {2^3} = 2.5\\9x - 8 = 10\\9x = 18\\x = 2\end{array}\)

Vậy \(x = 2\)

b)

\(\begin{array}{l}\left[ {{3^4} - \left( {{8^2} + 14} \right):13} \right]x = {5^3} + {10^2}\\\left[ {81 - \left( {64 + 14} \right):13} \right]x = 125 + 100\\\left[ {81 - 78:13} \right]x = 125 + 100\\\left[ {81 - 6} \right]x = 225\\75x = 225\\x = 3\end{array}\)

Vậy \(x = 3\)

mik cần gấp

a) \(18-\left(2x+5\right)=9\)

\(2x+5=18-9\)

\(2x+5=9\)

\(2x=9-5\)

\(2x=4\)

\(x=2\)

a) \(18-\left(2x+5\right)=9\)

\(\Rightarrow2x+5=18-9=9\)

\(\Rightarrow2x=9-5=4\Rightarrow x=4:2=2\)

b) \(23x-4=32\Rightarrow23x=32+4=36\Rightarrow x=\dfrac{36}{23}\)

c) \(\left(3x+2\right)^2=64\)

\(\Rightarrow\left[{}\begin{matrix}3x+2=8\\3x+2=-8\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{10}{3}\end{matrix}\right.\)

d) \(x\left(2x-12\right)=0\Rightarrow6x\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

2x+5=18-9

2x+5=9

2x=9-5

2x=4

  x=4:2

  x=2

c)3x+2=64:2

   3x+2=32

    3x=32-2

    3x=30

     x=30:3

     x=10

Mình cần gấp!!!!

Bài 1: 

a: x/33=4

nên x=4x33=132

b: 5/x=1/2

=>5/x=5/10

hay x=10

Bài 2:

Gọi tử là x

Mẫu là 180-x

Theo đề, ta có: \(\dfrac{x}{180-x}=\dfrac{4}{5}\)

=>5x=720-4x

=>9x=720

hay x=80

Vậy: Phân số cần tìm là 80/100

`a, 2/3 +3/4 = (8+9)/12=17/12.`

`1 1/3+4/5 = 4/3 + 4/5 = (20+12)/15=32/15`.

`=> x=2.`

`b, 5/6-1/4=(20-6)/24=7/12`.

`2 1/3-2/5= 7/3-2/5 = (35-6)/15=29/15`.

`=> x=1`.

a) \(\dfrac{2}{3}+\dfrac{3}{4}=\dfrac{8+9}{12}=\dfrac{17}{12}\)

-> 1 1/3 + 4/5 = 4/3 + 4/5 =  20+12/15 = 32/15

vậy x có thể = 14/14 = 1 (x thuộc N)

Bài 2: 

a: Ta có: \(2^{x+1}\cdot3^y=12^x\)

\(\Leftrightarrow2^{x+1}\cdot3^y=2^{2x}\cdot3^x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=2x\\x=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

a) 2(2x - 8) = 24

2x - 8 = 12

2x = 20

x = 10

b)3.(2^x - 26) = 18

2^x - 26 = 6

2^x = 32

x = 5

Bài 1: 

a: BCNN(10;12)=60

b: BCNN(24;10)=120

c: BCNN(4;14;26)=364

d: BCNN(6;8;10)=120