\(\dfrac{2x+19}{21}-\dfrac{2x+17}{23}=\dfrac{2x+7}{33}-\dfrac{2x+5}{35}\)

\(\Rightarrow\dfrac{2x+19}{21}-\dfrac{2x+17}{23}-\dfrac{2x+7}{33}+\dfrac{2x+5}{35}=0\)

\(\Rightarrow\left(\dfrac{2x+19}{21}+1\right)-\left(\dfrac{2x+17}{23}+1\right)-\left(\dfrac{2x+7}{33}+1\right)+\left(\dfrac{2x+5}{35}+1\right)=0\)

\(\Rightarrow\dfrac{2x+40}{21}-\dfrac{2x+40}{23}-\dfrac{2x+40}{33}+\dfrac{2x+40}{35}=0\)

\(\Rightarrow\left(2x+40\right)\left(\dfrac{1}{21}-\dfrac{1}{23}-\dfrac{1}{33}+\dfrac{1}{35}\right)=0\)

\(\Rightarrow2x+40=0\Rightarrow x=-20\)( do \(\dfrac{1}{21}-\dfrac{1}{23}-\dfrac{1}{33}+\dfrac{1}{35}>0\))

Ta có: \(\dfrac{2x+19}{21}-\dfrac{2x+17}{23}=\dfrac{2x+7}{33}-\dfrac{2x+5}{35}\)

\(\Leftrightarrow\left(2x+40\right)\left(\dfrac{1}{21}-\dfrac{1}{23}-\dfrac{1}{33}+\dfrac{1}{35}\right)=0\)

\(\Leftrightarrow2x+40=0\)

hay x=-20

a) \(6x+15\times8=12\times\left(19-x\right)\)

\(6x+120=228-12x\)

\(6x+120-228+12x=0\)

\(18x-108=0\)

\(18x=108\)

\(x=6\)

b) \(160-\left(35\div x+3\right)\times15=15\)

\(160-\left(35\div x+3\right)=1\)

\(35\div x+3=159\)

\(35\div x=156\)

\(x=\dfrac{35}{156}\)

c) \(2x-\left(1309\div11-19\right)-2=0\)

\(2x-1309\div11-19=2\)

\(2x-119-19=2\)

\(2x-119=21\)

\(2x=140\)

\(x=70\)

d) \(\left(x-7\right)\times\left(2x-16\right)=0\)

\(x-7=0;2x-16=0\)

\(x=7;2x=16\)

\(x=7;x=8\)

Bài giải chi tiết đây em nhé:

\(\dfrac{1}{3}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{63}\)+...+ \(\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}\) = \(\dfrac{9}{19}\)

\(\dfrac{1}{2}\)(\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+ \(\dfrac{2}{7.9}\)+...+ \(\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\)) = \(\dfrac{9}{19}\)

\(\dfrac{1}{2}\)( \(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+ \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) +... + \(\dfrac{1}{2x-1}-\dfrac{1}{2x+1}\)) = \(\dfrac{9}{19}\)

 \(\dfrac{1}{2}\) ( 1 - \(\dfrac{1}{2x+1}\)) = \(\dfrac{9}{19}\)

      1   - \(\dfrac{1}{2x+1}\) = \(\dfrac{9}{19}\) : \(\dfrac{1}{2}\)

       1  -  \(\dfrac{1}{2x+1}\) = \(\dfrac{18}{19}\)

               \(\dfrac{1}{2x+1}\) = \(1-\dfrac{18}{19}\)

                \(\dfrac{1}{2x+1}\) = \(\dfrac{1}{19}\)

                 \(2x+1\)  = 19

                 2\(x\)        = 19 - 1

                 2\(x\)       = 18

                    \(x\)      = 18: 2

                     \(x\)     = 9

a)|2x+19|=1

<=>2x+19=-1 or 2x+19=1

<=>2x=-20 or 2x=-18

<=>x=-10 or x=-9

b)|x-35|=6

<=>x-35=-6 or x-35=6

<=>x=29 or x=41

a,Ta có I 2x+19 I=1

         => 2x+19=-1,1

Nếu 2x+19 =1

   =>2x=20

   =>x=10

Nếu 2x+19=-1

  =>2x=-18

  =>x=-9

Vậy x=-9,10

b, Tương tự câu a. Tự làm đi. Cố lê!!!!!!!!!!

a) 42 - x = 5

x = 42 - 5

x = 37

b) 15.(x + 4) = 75

x + 4 = 75 : 15

x + 4 = 5

x = 5 - 4

x = 1

d) x + 18 = 12

x = 12 - 18

x = -6

e) 2x - 15 = -19

2x = (-19) + 15

2x = -4

x = -4 : 2

x = -2

f) 36 : (x^3 - 12) = -3

x^3 - 12 = 36 : (-3)

x^3 - 12 = -12

x^3 = 0

=> x = 0

g)x + 15 = 35

x = 35 - 15

x = 20

h) 2x - 13 = 3^2

2x - 13 = 9

2x = 9 + 13

2x = 22

x = 22 : 2

x = 11

a) 42-x=5

         x=42-5

         x= 37

b)15*(x+4)=75

        x+4=75:15

        x+4=5

        x    =5-4

       x     = 1