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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) \(3^7-x=1^4-\left(-3^5\right)\) 

\(2187-x=1-\left(-243\right)\) 

\(2187-x=244\) 

            \(x=2187-244\) 

            \(x=1943\)

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

a: x=67

b: x=44

c: x=-2 hoặc x=5

a: x=67

b: x=44

c: x=-2 hoặc x=5

a) x+7=-12

        x=(-12)-7

        x=-19

b)x-15=-21

        x=(-21)+15

        x=-6

c)13-x=20

        x=13-20

        x=-7

d)17-(2+x)=3

      x=17-3

      x=14

      x=14-2

      x=12

a,x+7=-12

=>x= -12-7

=>x= -19

b,x-15= -21

=>x= -21+15

=>x= -6

c,13-x=20

=>x=13-20

=>x= -7

d, 17-(2+x)=3

=>2+x=17-3

=>2+x=14

=>x=14-2

=>x=12

a) \(x+7=-12\) 

    \(x=-12-7\) 

    \(x=-19\) 

b) \(x-15=-21\)  

    \(x=-21+15\) 

    \(x=-6\)  

c) \(13-x=20\) 

            \(x=13-20\)  

            \(x=-7\) 

d) \(17-\left(2+x\right)=3\)  

             \(2+x=17-3\)  

             \(2+x=14\)   

                   \(x=14-2\)  

                   \(x=12\)

https://m.youtube.com/watch?v=TczQYf7-Gqk

a) Ta có: 12-5x=37

\(\Leftrightarrow5x=-25\)

hay x=-5

Vậy: x=-5

b) Ta có: 7-3|x-2|=-11

\(\Leftrightarrow3\left|x-2\right|=18\)

\(\Leftrightarrow\left|x-2\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)

Vậy: \(x\in\left\{8;-4\right\}\)

c) Ta có: \(x+\dfrac{2}{8}=-\dfrac{15}{4}\)

\(\Leftrightarrow x=\dfrac{-15}{4}-\dfrac{2}{8}=\dfrac{-15}{4}-\dfrac{1}{4}\)

hay x=-4

Vậy: x=-4

a, \(\Leftrightarrow5x=12-37=-25\)

\(\Leftrightarrow x=-\dfrac{25}{5}=-5\)

Vậy ...

b, \(\Leftrightarrow3\left|x-2\right|=7+11=18\)

\(\Leftrightarrow\left|x-2\right|=\dfrac{18}{3}=6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)

Vậy ...

c, \(\Leftrightarrow x=-\dfrac{15}{4}-\dfrac{2}{8}=-4\)

Vậy ..

a)-5; b)8 hoặc -4;c)-4 nhé. Bn tham khảo nhé