a/ \(x\left(x+7\right)=0\) \(\Rightarrow\left[\begin{matrix}x=0\\x+7=0\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}x=0\\x=-7\end{matrix}\right.\)

b/ \(\left(-x+5\right)\left(3-x\right)=0\) \(\Rightarrow\left[\begin{matrix}-x+5=0\\3-x=0\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}x=5\\x=3\end{matrix}\right.\)

c/ \(\left|x-1\right|=3\) \(\Rightarrow\left[\begin{matrix}x-1=3\\1-x=3\end{matrix}\right.\) \(\Rightarrow\left[\begin{matrix}x=4\\x=-2\end{matrix}\right.\)

d/ \(-13\left|x\right|=-26\) \(\Rightarrow\left|x\right|=2\) \(\Rightarrow x=\pm2\)

e/ \(x.x-8=-2.\left(-13\right)-\left(-2\right)\)

\(\Rightarrow x^2=36\) \(\Rightarrow\left|x\right|=6\) \(\Rightarrow x=\pm6\)

a, ( x + 1 ) + ( x + 2 ) + ... + ( x + 199 ) = 0

x + 1 + x + 2 + ... + x + 199 = 0

( x + x + ... + x ) + ( 1 + 2 + ... + 199 ) = 0 

199x + 19900 = 0

199x = 0 - 19900

199x = -19900

x = -19900 : 199

x = -100

Vậy ...

b, ( x - 30 ) + ( x - 29 ) + ( x - 28 ) = 11

x - 30 + x - 29 + x - 28 = 11

( x + x + x ) - ( 30 + 29 + 28 ) = 11

3x -  87 = 11

3x = 11 + 87 

3x = 98

x = \(\frac{98}{3}\)

Vậy ... 

123 - 3 ( x + 4 ) = 23

3 ( x + 4 ) = 123 - 23

3 ( x + 4 ) = 100

x + 4 = \(\frac{100}{3}\)

x = \(\frac{100}{3}-4\)

x = \(\frac{88}{3}\)

Vậy ...

-16 + 23 + x = -16

          7 + x = -16 

               x = -16 - 7

               x = -23

2x + 35 = -15

        2x = -15 - 35

         2x = -50

           x = -25

-13 x |x | = -26

         x = -26 : (-13)

          x = 2

Vậy x = 2 hoặc -2

|2x - 5| = 13

      2x = 13 + 5

       2x = 18

         x = 9

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

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

2)2x+35=-15

2x      =-15-35

2x      = -50

x        =-25

1)-16+23+x=-16

7+x          =-16

x             =-16-7

x             =-23

3)-13 * lxl =-26

lxl         =  -26:-13

lxl           =2

<=>x=-2 hoac x=2

5) (x-3).(x+2) =0

<=>\(\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\) <=>\(\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

a) \(x-\dfrac{2}{3}=\dfrac{3}{8}\Rightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{25}{24}\)

b) \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\Rightarrow x-\dfrac{3}{4}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}+\dfrac{3}{4}=1\)

c) \(\dfrac{3}{2}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{2}-\dfrac{4}{5}=\dfrac{7}{10}\)

\(\Rightarrow x=\dfrac{7}{10}-\dfrac{1}{2}=\dfrac{1}{5}\)

d) \(\left|x-2\right|-1=0\Rightarrow\left|x-2\right|=1\)

\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

a: Ta có: \(x-\dfrac{2}{3}=\dfrac{3}{8}\)

\(\Leftrightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{9}{24}+\dfrac{16}{24}=\dfrac{25}{24}\)

b: Ta có: \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\)

\(\Leftrightarrow x-\dfrac{3}{4}=\dfrac{13}{10}\cdot\dfrac{5}{26}=\dfrac{1}{4}\)

hay x=1

a) x - 26 : 13 = 2017

             x - 2 = 2017

                  x = 2017 + 2

                  x = 2019

b) \(\frac{2}{3}+\frac{1}{3}.\left(x+28\right)=18\)

                \(\frac{1}{3}.\left(x+28\right)=18-\frac{2}{3}\)

                \(\frac{1}{3}.\left(x+28\right)=\frac{54}{3}-\frac{2}{3}\)

                \(\frac{1}{3}.\left(x+28\right)=\frac{52}{3}\)

                             \(x+28=\frac{52}{3}:\frac{1}{3}\)

                             \(x+28=52\)

                                         \(x=52-28\)

                                         \(x=24\) 

c) \(\frac{7}{8}.\left(x-27\right)=\frac{9}{8}-0,125\)

   \(\frac{7}{8}.\left(x-27\right)=1\)

                \(x-27=1:\frac{7}{8}\)

                \(x-27=\frac{8}{7}\)

                            \(x=\frac{8}{7}+27\)

                            \(x=\frac{8}{7}+\frac{189}{7}\)

                           \(x=\frac{197}{7}\)

a.17-(-x(-x(-x)))=16

17+x-x+x=16

17+x=16

x=16-17

x=-1

b.26-|x+19|=-13

|x+19|=26--13

|x+19|=26+13

|x+19|=39

\(\Rightarrow\)\(\orbr{\begin{cases}x+19=39\\x+19=-39\end{cases}}\)

\(\Rightarrow\)\(\orbr{\begin{cases}x=20\\x=-58\end{cases}}\)

Ta có: \(0,\left(3\right)+\frac{31}{3}+0,4\left(2\right)=\frac{3}{9}+\frac{31}{3}+\frac{42-4}{90}=\frac{1}{3}+\frac{31}{3}+\frac{19}{45}=\frac{32}{3}+\frac{19}{45}=\frac{499}{45}.\)

\(\frac{4}{9}+0,\left(13\right)=\frac{4}{9}+\frac{13}{99}=\frac{44}{99}+\frac{13}{99}=\frac{57}{99}=\frac{19}{33}\)

\(0,\left(37\right).x\Rightarrow\frac{37}{99}.x=1\)

\(\Rightarrow x=1:\frac{37}{99}=\frac{99}{37}\)

\(0,\left(26\right).x=1,2\left(31\right)\)

\(\Rightarrow\frac{26}{99}.x=\frac{1219}{990}\)

\(\Rightarrow x=\frac{1219}{990}:\frac{26}{99}=\frac{1219}{260}\)