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

\(45⋮x\)

\(\Rightarrow x\inƯ\left(45\right)=\left\{1;3;5;9;15;45\right\}\)

Vậy \(x\in\left\{1;3;5;9;15;45\right\}\)

Vì 45 chia hết cho x => x thuộc vào Ư(45)

Ư(45) thuộc vào tập hợp 1,3,5,9,15,45

=>x thuộc 1,3,5,9,15,45

Thông cảm nha mình tự làm nên ko viết được kí hiệu

B3: 

a) x = 0

b) 9

B4 :

a) x = 1,2,3

b) x = 0,1,2

B5:

0,21; 0,215; 0,22

a) \(6x-5=613\)

\(\Leftrightarrow6x=618\)

\(\Leftrightarrow x=103\)

b) \(74\left(x-3\right)=0\)

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\)

\(a,\Rightarrow6x=613+5=618\\ \Rightarrow x=103\\ b,\Rightarrow x-3=0\Rightarrow x=3\)

a,6.x-5=613

   6x    =613+5=618

     x     =103

 Vậy........

b,74.(x-3)=0

         x-3  =0:74=0

         x     = 0+3=3

   Vậy .....

a: =>x>5

a, 35 - \(x\) < 35 - 5

    35 - \(x\) < 30

           \(x\) < 35 - 30

           \(x\) < 5

\(x\) = 1; 2; 3; 4; 5

b, \(x\) - 10 < 35 - 10

    \(x\) - 10 < 15

   \(x\)         < 15 + 10

   \(x\)        < 25

   \(x\) = 0; 1; 2; 3; 4; 5; ....;24

c, \(x\) - 10 < 45

    \(x\)        < 45 + 10

    \(x\)         < 55

   \(x\) = 0; 1; 2; 3; 4;...;54

\(a,2448:\left[119-\left(x-6\right)\right]=24\)

\(119-\left(x-6\right)=2448:24\)

\(119-\left(x-6\right)=102\)

\(x-6=119-102\)

\(x-6=17\)

\(x=17+6\)

\(\Rightarrow x=23\)

\(b,2016-100\times\left(x+11\right)=128:8\)

\(2016-100\times\left(x+11\right)=16\)

\(100\times\left(x+11\right)=2016-16\)

\(100\times\left(x+11\right)=2000\)

\(x+11=2000:100\)

\(x+11=20\)

\(x=20-11\)

\(\Rightarrow x=9\)

\(a,2448:\left[119-\left(x-6\right)\right]=24\\ 119-\left(x-6\right)=2448:24\\ 119-\left(x-6\right)=102\\ x-6=119-102\\ x-6=17\\ x=17+6\\ x=23\\ b,2016-100\times\left(x+11\right)=128:8\\ 2016-100\times\left(x+11\right)=16\\ 100\times\left(x+11\right)=2016-16\\ 100\times\left(x+11\right)=2000\\ x+11=2000:100\\ x+11=20\\ x=20-11\\ x=9\)

\(a,9x-8=838\)

\(9x=838+8\)

\(9x=846\)

\(x=846:9=94\)

\(b,\left(x+30\right)-75=125\)

\(x+30=125+75\)

\(x+30=200\)

\(x=200-30=170\)

\(c,x-72:36=18\)

\(x-2=18\)

\(x=18+2=20\)

#Urushi

a) \(5\times x-123=12\)

\(\Rightarrow5\times x=135\)

\(\Rightarrow x=27\)

b) \(x+3x+5x+7x=96\)

\(\Rightarrow16x=96\)

\(\Rightarrow x=6\)

a) \(5\times x-123=12\)

\(5x=12+123\)

\(5x=135\)

\(x=135:5\)

\(x=27\)

________

b) \(x+3x+5x+7x=96\)

\(x\left(1+3+5+7\right)=96\)

\(x.16=96\)

\(x=96:16\)

\(x=6\)

a)

\(5\times x-123=12\\\Rightarrow 5\times x=12+123\\\Rightarrow 5\times x=135\\\Rightarrow x=135:5\\\Rightarrow x=27\\---\)

b) 

\(x+3x+5x+7x=96\\\Rightarrow16x=96\\\Rightarrow x=96:16\\\Rightarrow x=6\)

a) 0,75 < x < 3,25

x = 1 hoặc 2 hoặc 3

b) x < 2,02

x = 2

a) 1;2;3
b) 1;2

Bài 1:

a: Ta có: \(48751-\left(10425+y\right)=3828:12\)

\(\Leftrightarrow y+10425=48751-319=48432\)

hay y=38007

b: Ta có: \(\left(2367-y\right)-\left(2^{10}-7\right)=15^2-20\)

\(\Leftrightarrow2367-y=1222\)

hay y=1145

Bài 2: 

Ta có: \(8\cdot6+288:\left(x-3\right)^2=50\)

\(\Leftrightarrow288:\left(x-3\right)^2=2\)

\(\Leftrightarrow\left(x-3\right)^2=144\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)