a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)

\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{-5}{12}\)

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

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{5}\)

c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)

\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)

\(\Rightarrow x=\dfrac{-77}{120}\)

d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)

\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)

\(\Rightarrow x=\dfrac{-7}{20}\)

e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)

\(\Rightarrow x=\dfrac{-59}{105}\)

g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

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

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

\(\Rightarrow x=\dfrac{-13}{12}\)

1: =>(x+2018)(6x-3)=0

=>x+2018=0 hoặc 6x-3=0

=>x=1/2 hoặc x=-2018

2: x(x-11)+3(11-x)=0

=>(x-11)(x-3)=0

=>x=11 hoặc x=3

4: =>(x+5)(2x-4)=0

=>2x-4=0 hoặc x+5=0

=>x=2 hoặc x=-5

3: =>(x-3)(x+2)=0

=>x=3 hoặc x=-2

Bài 1:

\(6x\left(x+2018\right)-3\left(x+2018\right)=0\)

\(\Leftrightarrow\left(x+2018\right)\left(6x-3\right)=0\)

\(\Leftrightarrow3\left(x+2018\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2018\\2x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2018\\x=\dfrac{1}{2}\end{matrix}\right.\)

Bài 2:

\(x\left(x-11\right)+3\left(11-x\right)=0\)

\(\Leftrightarrow x\left(x-11\right)-3\left(x-11\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=11\end{matrix}\right.\)

Câu 3:

\(x\left(x-3\right)-2\left(3-x\right)=0\)

\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Câu 4:

\(2x\left(x+5\right)-4\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\2x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

|7 - \(\dfrac{3}{4}\)\(x\)| - \(\dfrac{3}{2}\) = \(\dfrac{1}{\dfrac{1}{2}}\)

|7 - \(\dfrac{3}{4}x\)|  - \(\dfrac{3}{2}\) = 2

|7 - \(\dfrac{3}{4}\)\(x\)| = 2 + \(\dfrac{3}{2}\)

|7 - \(\dfrac{3}{4}x\)| = \(\dfrac{7}{2}\)

\(\left[{}\begin{matrix}7-\dfrac{3}{4}x=\dfrac{7}{2}\\7-\dfrac{3}{4}x=-\dfrac{7}{2}\end{matrix}\right.\) 

\(\left[{}\begin{matrix}\dfrac{3}{4}x=7-\dfrac{7}{2}\\\dfrac{3}{4}=7+\dfrac{7}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{7}{2}\\\dfrac{3}{4}x=\dfrac{21}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{14}{3}\\x=14\end{matrix}\right.\)

 5  - |\(x-3\)| = 5

       |\(x-3\)| = 5 - 5

      |\(x-3\)| = 0

      \(x-3\) = 0 

      \(x\) = 3

|\(\dfrac{7}{2}\) - 42| + 14 = 14 ( vô lý xem lại đề bài nhé em)

\(\dfrac{2}{3}\) - |\(x+2\)| = 2

     |\(x+2\)| = \(\dfrac{2}{3}\) - 2

    |\(x+2\)| = - \(\dfrac{4}{3}\)  

   |\(x+2\)| ≥ 0 ∀ \(x\)

Vậy \(x\in\) \(\varnothing\)

a) \(\dfrac{1}{2}-\left(x+\dfrac{1}{3}\right)=\dfrac{5}{6}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{5}{6}\)

\(\Rightarrow x+\dfrac{1}{3}=\dfrac{-1}{3}\)

\(\Rightarrow x=\dfrac{-1}{3}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{-2}{3}\)

b)\(\dfrac{3}{4}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\)

\(\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)

\(\Rightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)

\(\Rightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{-11}{20}\)

c) \(\dfrac{3}{35}-\left(\dfrac{3}{5}+x\right)=\dfrac{2}{7}\)

\(\Rightarrow\dfrac{3}{5}+x=\dfrac{3}{35}-\dfrac{2}{7}\)

\(\Rightarrow\dfrac{3}{5}+x=\dfrac{-1}{5}\)

\(\Rightarrow x=\dfrac{-1}{5}-\dfrac{3}{5}\)

\(\Rightarrow x=\dfrac{-4}{5}\)

d)\(\dfrac{2}{3}.x=\dfrac{4}{27}\)

\(\Rightarrow x=\dfrac{4}{27}:\dfrac{2}{3}\)

\(\Rightarrow x=\dfrac{2}{9}\)

e) \(\dfrac{-3}{5}.x=\dfrac{21}{10}\)

\(\Rightarrow x=\dfrac{21}{10}:\dfrac{-3}{5}\)

\(\Rightarrow x=\dfrac{-7}{2}\)

\(\frac{3}{x-5}=-\frac{4}{x+2}\)

=> 3 ( x + 2 ) = 4 ( x - 5 ) 

=> 3x + 6 = 4x - 20

=> 3x - 4x = - 6 - 20

=> - 1x = - 26

=> x = 26

1) 3/x-5=-4/x+2

     3x+6=-4x+20

     7x .  =14

      x  .   =2

2) x+3/-4=-9/x+3

   (x+3)^2=36

  Ta có hai trường hợp:

*x+3=6=)x=3

*x+3=-6=)x=-9

3) (34/5-2x). 11/3=55/7

     34/5-2x .          =15/7

      75-30x.            =238

           30x.            = -163

               x.            =-163/30

a) ( x - 1 )( x² + x + 1 ) + x( x + 2 )( 2 - x ) = 5

<=> x³ - 1 - x( x + 2 )( x - 2 ) = 5 

<=> x³ - 1 - x( x² - 4 ) = 5

<=> x³ - 1 - x³ + 4x = 5

<=> 4x - 1 = 5

<=> 4x = 6

<=> x = 6/4 = 3/2

b) 5x( x - 3 )² - 5( x - 1 )³ + 15( x + 4 )( x - 4 ) = 5

<=> 5x( x² - 6x + 9 ) - 5( x³ - 3x² + 3x - 1 ) + 15( x² - 16 ) = 5

<=> 5x³ - 30x² + 45x - 5x³ + 15x² - 15x + 5 + 15x² - 240 = 5

<=> 30x - 235 = 5

<=> 30x = 240

<=> x = 8

a,\(\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\)

\(< =>x^3-1+x\left(4-x^2\right)=5\)

\(< =>x^3-1+4x-x^3=5\)

\(< =>4x-1-5=0< =>4x-6=0< =>x=\frac{3}{2}\)

b, \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+4\right)\left(x-4\right)=5\)

\(< =>5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15\left(x^2-16\right)=5\)

\(< =>5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-240=5\)

\(< =>\left(5x^3-5x^3\right)+\left(15x^2+15x^2-30x^2\right)+\left(45x-15x\right)+5-240=5\)

\(< =>30x-240=5-5=0< =>x=\frac{24}{3}=8\)

a, \(\left(x-1\right)\left(x^2+x+1\right)+x\left(x+2\right)\left(2-x\right)=5\)

\(\Leftrightarrow x^3-1+4x-x^3=5\Leftrightarrow-6+4x=0\Leftrightarrow x=\frac{3}{2}\)

b, \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+4\right)\left(x-4\right)=5\)

\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-240=5\)

\(\Leftrightarrow30x-240=5\Leftrightarrow x=8\)