đk : x khác -2 ; -1/2 

\(\Rightarrow6x+3=5x+10\Leftrightarrow x=7\left(tmđk\right)\)

ĐKXĐ:\(x\ne-2,x\ne-\dfrac{1}{2}\)

\(\dfrac{3}{x+2}=\dfrac{5}{2x+1}\\ \Leftrightarrow3\left(2x+1\right)=5\left(x+2\right)\\ \Leftrightarrow6x+3=5x+10\\ \Leftrightarrow x=7\left(tm\right)\)

\(\Rightarrow3\cdot\left(2x+1\right)=\left(x+2\right)\cdot5\\ \Rightarrow6x+3=5x+10\\ \Rightarrow6x-5x=10-3\\ \Rightarrow x=7\)

a, x= -9/2

b, x= 2

kb nha

a) \(\left|x+5\right|-x+23=27\)

\(\Leftrightarrow\left|x+5\right|-x+23-27=0\)

\(\Leftrightarrow\left|x+5\right|-x+\left(23-27\right)=0\)

\(\Leftrightarrow\left|x+5\right|-x+\left(-4\right)=0\)

\(\Leftrightarrow\left|x+5\right|-x-4=0\)

\(\Leftrightarrow\left|x+5\right|-\left(x+4\right)=0\)

\(\Leftrightarrow\left|x+5\right|=x+4\)

Xét trường hợp 1:

\(x+5=x+4\)

\(\Leftrightarrow x+5-\left(x+4\right)=0\)

\(\Leftrightarrow x+5-x-4=0\)

\(\Leftrightarrow1=0\left(loai\right)\)

Xét trường hợp 2:

\(x+5=-\left(x+4\right)\)

\(\Leftrightarrow x+5-\left[-\left(x+4\right)\right]=0\)

\(\Leftrightarrow x+5+x+4=0\)

\(\Leftrightarrow2x+9=0\)

\(\Leftrightarrow2x=-9\)

\(\Leftrightarrow x=\dfrac{-9}{2}\left(thoa\right)\)

Vậy \(x=\dfrac{-9}{2}\)

a,\(x+3=15\)

\(=>x=15-3=12\)

b,\(x-7+3=25\)

\(=>x=25+7-3=29\)

c,\(x-3-6=-4\)

\(=>x=-4+6+3=5\)

d,\(26-x+9=-13\)

\(=>26-x=-13-9=-22\)

\(=>26+22=x\)

\(=>x=48\)

x+3=15

    x=15-3

    x=12

Vậy x=12

x-7+3=25

x-7=25-3

x-7=22

   x=22+7

   x=29

Vậy x=29

x-3-6=-4

x-3=-4+6

x-3=2

   x=2+3

   x=5

Vậy x=5

26-x+9=-13

26-x=-13-9

26-x=22

     x=26-22

     x=4

Vậy x=4

\(x+3=15\Leftrightarrow x=15-3=12\)

\(x-7+3=25\Leftrightarrow x-7=25-3=22\Leftrightarrow x=22+7=29\)

\(x-3-6=-4\Leftrightarrow x-3=\left(-4\right)+6=2\Leftrightarrow x=2+3=5\)

\(26-x+9=-13\Leftrightarrow26-x=\left(-13\right)-9=-22\Leftrightarrow x=26-\left(-22\right)=48\)

\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Rightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

\(\Rightarrow x+10=0\) vì \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)

\(\Rightarrow x=-10\)

=>\(\frac{x+1}{9}\)+1+\(\frac{x+2}{8}\)+1=\(\frac{x+3}{7}\)+1+\(\frac{x+4}{6}\)+1

=>\(\frac{x+10}{9}\)+ \(\frac{x+10}{8}\)- \(\frac{x+10}{7}\)- \(\frac{x+10}{6}\)= 0

=>(x+10)x(\(\frac{1}{9}\)+ \(\frac{1}{8}\)+ \(\frac{1}{7}\)+\(\frac{1}{6}\))=0

=>x+10=0

=>x= -10

\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

Vì \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)nên x +10  = 0

x = -10.

Bài 1

a) x ⋮ 6 ⇒ x ∈ B(6) = {0; 6; 12; 18; 24; ...}

Mà 10 < x < 18 nên x = 12

b) 24 ⋮ x ⇒ x ∈ Ư(24) = {1; 2; 3; 4; 6; 8; 12; 24}

Mà x > 4

⇒ x ∈ {6; 8; 12; 24}

c) x ⋮ 10 ⇒ x ∈ B(10) = {0; 10; 20; 30; 40;...}  (1)

Lại có 45 ⋮ x ⇒ x ∈ Ư(45) = {1; 3; 5; 9; 15; 45}  (2)

Từ (1) và (2) ⇒ không tìm được x thỏa mãn đề bài

Bài 2

a) *) (60 + x) ⋮ 5

Mà 60 ⋮ 5

⇒ x ⋮ 5

⇒ x = 5k (k )

*) (72 - x) ⋮ 5

72 chia 5 dư 2

⇒ x chia 5 dư 3

⇒ x = 5k + 3 (k ∈ ℕ)

b) Gọi a, a + 1, a + 2 là ba số tự nhiên liên tiếp (a ∈ ℕ)

Ta có:

a + a + 1 + a + 2

= 3a + 3

= 3(a + 1) ⋮ 3

Vậy tổng ba số tự nhiên liên tiếp chia hết cho 3

a,123-5(x+4)=28

(=)5(x+4)=85

(=) x+4=17

(=) x=13

b,10+2x=2.\(\left(3^2-1\right)\)

(=) 10+2x=16

(=) 2x = 6 

(=)x=3

A, 123-5.(x+4)=38

            5.(x+4)=123-38

           5.(x+4) =85

               (x+4)=85:5

               (x+4)=17

                x      =17-4

                x      =13

B, 10+2x=2(\(3^2-1\))

     10+2x=2.8

     10+2x=16

           2x=16-10

           2x=6

             x=6:2

            x  =3

a, 123 - 5(x + 4) = 38

=>5(x + 4) = 85

=> x + 4 = 17

=> x = 13

vậy_

10 + 2x = 2(3² - 1)

=> 10 + 2x = 2.8

=> 10 + 2x = 16

=> 2x = 6

=> x = 3

vậy_

a) |x| + |-5| = |-37|

<=> |x| + 5 = 37

<=> |x| = 37 - 5 = 32

=> x \(\in\) {32 ; -32}

b)|-6| . |x| = |54|

<=> 6 . |x| = 54

|x| = 54 : 6 = 9

=> x \(\in\){9;-9}

c) |x| > 21

Có |x| \(\ge\) 0 > 21

=> |x| \(\in\) { 22 ; 23 ; 24 ; 25 ; ....}

=> x \(\in\) { 22; -22 ; 23; -23; 24; -24; 25; -25; ....}