a: Ta có: \(7x+25=144\)

\(\Leftrightarrow7x=119\)

hay x=17

b: Ta có: \(33-12x=9\)

\(\Leftrightarrow12x=24\)

hay x=2

c: Ta có: \(128-3\left(x+4\right)=23\)

\(\Leftrightarrow3\left(x+4\right)=105\)

\(\Leftrightarrow x+4=35\)

hay x=31

d: Ta có: \(71+\left(726-3x\right)\cdot5=2246\)

\(\Leftrightarrow5\left(726-3x\right)=2175\)

\(\Leftrightarrow726-3x=435\)

\(\Leftrightarrow3x=291\)

hay x=97

e: Ta có: \(720:\left[41-\left(2x+5\right)\right]=40\)

\(\Leftrightarrow41-\left(2x+5\right)=18\)

\(\Leftrightarrow2x+5=23\)

\(\Leftrightarrow2x=18\)

hay x=9

Bn cần bài nào vậy

f: Ta có: \(10-4x+120:8=16+1\)

\(\Leftrightarrow4x=17-25=-8\)

hay x=-2

g: Ta có: \(x+9x+7x+5x=2244\)

\(\Leftrightarrow22x=2244\)

hay x=102

h: Ta có: \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)

\(\Leftrightarrow100x+5050=5750\)

\(\Leftrightarrow100x=700\)

hay x=7

a: =>41-(2x-5)=720:40=18

=>2x-5=23

=>2x=28

=>x=14

b: =>100x+5050=5750

=>100x=700

=>x=7

a: =>41-(2x-5)=720:40=18

=>2x-5=23

=>2x=28

=>x=14

b: =>100x+5050=5750

=>100x=700

=>x=7

a)128-3(x+4)=23 

    125.(x+4)=23

             ( x +4)=23 : 125

             x+4 = 0,184

                x = 0,184 -4

                  x = -3 .816

theo mềnh nhóe ( làm từng câu )

c: =>12x-64=32

=>12x=96

hay x=8

g: =>3x-16=14

hay x=10

e: =>5(x+4)=85

=>x+4=17

hay x=13

a)720:[41-(2x-5)]=40

41-(2x-5)=720:40

41-(2x-5)=18

       2x-5=41-18

       2x-5=23

       2x=28

          x=14

\(2x=28\)

\(x=14\)

chúc bn học tốt.

b) x=7,a) x= -35

1.

a) (2x + 1)³ =343

\(\Leftrightarrow\left(2x+1\right)^3=7^3.\)

\(\Leftrightarrow2x+1=7\Rightarrow x=3\)

b) 720 [ 41 (2x - 5)] = 2³ . 5

\(\Leftrightarrow720\left[41\left(2x-5\right)\right]=40\)

\(\Leftrightarrow41\left(2x-5\right)=\frac{40}{720}=\frac{1}{18}\)

\(\Leftrightarrow2x-5=\frac{1}{738}\)

\(\Leftrightarrow2x=\frac{3691}{738}\Leftrightarrow x=\frac{3691}{1476}\)

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