a; \(\dfrac{93}{17}\): \(x\) + (- \(\dfrac{21}{17}\)) : \(x\) + \(\dfrac{22}{7}\): \(\dfrac{22}{3}\) = \(\dfrac{5}{14}\)

   \(\dfrac{94}{17}\) \(\times\) \(\dfrac{1}{x}\) - \(\dfrac{21}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)

    \(\dfrac{72}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)

     \(\dfrac{72}{17x}\)        = \(\dfrac{5}{14}\) - \(\dfrac{3}{7}\)

      \(\dfrac{72}{17x}\)      = - \(\dfrac{1}{14}\)

      17\(x\)       = 72.(-14)

       17\(x\)     = - 1008

         \(x\)       = - 1008 : 17

         \(x\)       = - \(\dfrac{1008}{17}\)

Vậy \(x\) \(=-\dfrac{1008}{17}\)

b; - \(\dfrac{32}{27}\) - (3\(x\) - \(\dfrac{7}{9}\))³ = - \(\dfrac{24}{27}\)

          - \(\dfrac{32}{27}\)  + \(\dfrac{24}{27}\) = (3\(x\) - \(\dfrac{7}{9}\))³ 

          (3\(x-\dfrac{7}{9}\))³ = - \(\dfrac{8}{27}\)

         (3\(x-\dfrac{7}{9}\))³ = (- \(\dfrac{2}{3}\))³

           3\(x-\dfrac{7}{9}\) = - \(\dfrac{2}{3}\)

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

             \(x\)        = \(\dfrac{1}{9}\) : 3

             \(x\)       = \(\dfrac{1}{27}\)

Vậy \(x=\dfrac{1}{27}\)

\(a)x+\frac{2}{3}=\frac{5}{6}\)

\(x=\frac{5}{6}-\frac{2}{3}\)

\(x=\frac{5}{6}-\frac{4}{6}=\frac{1}{6}\)

\(b)x-\frac{3}{5}=\frac{2}{7}+\frac{1}{5}\)

\(x=\frac{3}{5}-\frac{17}{35}\)

\(x=\frac{21}{35}-\frac{17}{35}\)

\(x=\frac{4}{35}\)

\(c)x\div\frac{9}{11}=\frac{5}{22}\)

\(x=\frac{5}{22}\times\frac{9}{11}\)

\(x=\frac{45}{242}\)

\(d)\frac{3}{5}\times x=\frac{3}{10}\)

\(x=\frac{3}{10}\div\frac{3}{5}\)

\(x=\frac{1}{2}\)

\(a)\)\(x+\frac{2}{3}=\frac{5}{6}\)

\(x=\frac{5}{6}-\frac{2}{3}\)

\(x=\frac{1}{6}\)

\(b)\)\(x-\frac{3}{5}=\frac{2}{7}+\frac{1}{5}\)

\(x-\frac{3}{5}=\frac{17}{35}\)

\(x=\frac{17}{35}+\frac{3}{5}\)

\(x=\frac{38}{35}\)

\(c)\)\(x\div\frac{9}{11}=\frac{5}{22}\)

\(x=\frac{5}{22}\times\frac{9}{11}\)

\(x=\frac{45}{242}\)

\(d)\)\(\frac{3}{5}\times x=\frac{3}{10}\)

\(x=\frac{3}{10}\div\frac{3}{5}\)

\(x=\frac{3}{10}\times\frac{5}{3}\)

\(x=\frac{1}{2}\)

\(\dfrac{\dfrac{22}{15}}{11-x}\) + \(\dfrac{2}{3}\) = \(\dfrac{7}{5}\)

\(\dfrac{\dfrac{22}{15}}{11-x}\) = \(\dfrac{7}{5}\) - \(\dfrac{2}{3}\)

\(\dfrac{\dfrac{22}{15}}{11-x}\) = \(\dfrac{11}{15}\) 

11 - \(x\) = \(\dfrac{22}{15}\) : \(\dfrac{11}{15}\)

11 - \(x\) = 2

        \(x\) = 11 - 2

         \(x\) = 9

a) x+3/22=-27/121.11/9

⇒ x+3/22=-3/11

⇒ x=-9/22

b) (3/2/20-2/1/20)x+2/7=3/5

⇒ (31/10-41/20)x=11/35

⇒ 21/20x=11/35

⇒x=44/147

c) |3/2x-5|-1/2=-1/4

⇒ |3/2x-5|=1/4

⇒ hoặc 3/2x-5=1/4⇒3/2x=21/4⇒x=7/2

hoặc 3/2x-5=-1/4⇒3/2x=19/4⇒x=19/6

a) x - 14 = 3x + 18

=> x - 3x = 18 + 14

=> -2x = 32

=> x = 32 : (-2)

=> x = -16

b) 2(x - 5) - 3(x - 4) = -6 + 15.(-3)

=> 2x - 10 - 3x + 12 = -6 - 45

=> -x + 2 = -51

=> -x = -51 - 2

=> -x = -53

=> x = 53

c) (x + 7)(x - 9) = 0

=> \(\orbr{\begin{cases}x+7=0\\x-9=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-7\\x=9\end{cases}}\)

Vậy ...

d) |2x - 5| - 7 = 22

=> |2x - 5| = 22 + 7

=> |2x - 5| = 29

=> \(\orbr{\begin{cases}2x-5=29\\2x-5=-29\end{cases}}\)

=> \(\orbr{\begin{cases}2x=34\\2x=-24\end{cases}}\)

=> \(\orbr{\begin{cases}x=17\\x=-12\end{cases}}\)

Vậy ...

\(x-14=3x+18\)

\(x-3x=14+18\)

\(-2x=32\)

\(-x=16\)

\(\Leftrightarrow x=-16\)

d) \(\left|2x-5\right|-7=22\)

\(\Rightarrow\left|2x-5\right|=22+7\)

\(\Rightarrow\left|2x-5\right|=29\)

\(\Rightarrow\orbr{\begin{cases}2x-5=29\\2x-5=-29\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}2x=34\\2x=-24\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=17\\x=-12\end{cases}}\)

Vậy x = 17 hoặc x = -12

d) |2x – 5| –7 = 22

| 2x -5 | = 22+7

| 2x -5 | = 29

TH1: 2x-5 = -29

2x = -29+5

2x= -24

x= -24:2

x= -12

TH2: 2x -5 =29

2x = 29+5

2x= 34

x= 34:2

x= 17 

Vậy...

a) x – 14 = 3x + 18

x -14 - 3x = 18

-2x -14 =18

-2x = 18-14

-2x = 4

x = 4:( -2) 

x= -2

Vậy...

b) 2 ( x – 5 ) – 3 ( x – 4 ) = -6 + 15.(- 3)

2 ( x – 5 ) – 3 ( x – 4 ) = -51

2x -10 - 3x -12 = -51

-x -  [ -10 - ( -12) ] = -51

-x - 2 = -51

-x = -51+2

-x = -49

Vậy...

Mk thử lại thấy nó sai sai nhưng ko biết sửa sao????

c) ( x + 7 ) ( x – 9) = 0 

=> x+ 7 = 0 hoặc x- 9 = 0

Nếu x+7 = 0 thì x = 0-7 => x = -7

Nếu x-9 = 0 thì x = 0+9=> x = 9

Vậy...