3x+17=12

=>3x=12-17

=>3x=-5

=>x=-5:3

=>x=-1,66

/x-1/=0

=>x-1=0

=>x   =0+1

=>x  =1

-13./x/=-26

=>/x/=-26:(-13)

=>/x/=2

=>x=-2 hoặc x=2

|x-1|=0

\(\Rightarrow\)x-1=0

x=0+1

x=1

Vậy x=1.

-13.|x|=26

|x|=-2

Mà |x|\(\ge\)0

\(\Rightarrow x\in\varnothing\)

Vậy \(x\in\varnothing\)

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#Huyền#

a, | x + 1 | = 0

=> x + 1 = 0

=> x = - 1

Vậy x = - 1

b,  - 13 . | x | = - 26

=>  | x| = 2

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

Vậy \(x\in\left\{2;-2\right\}\)

@@ Học toost

a) I x - 1 I=0 => x = 1

b) -13. I x I = -26

<=> IxI= -26 : (-13)=2

=> x= -2 hoặc x = 2

vậy x= -2 hoặc x=2

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)  \(\left|x-1\right|=0\) 

<=>\(x-1=0\)<=> \(x=1\)

vậy \(x=1\)

b) \(-13.\left|x\right|=-26\)

<=> \(\left|x\right|=2\)=> \(x=\pm2\)

vậy \(x=\pm2\)

c)  \(-16+23+x=-16\)

<=> \(7+x=-16\)

<=>\(x=-23\)

vậy \(x=-23\)

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

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

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

1,-16+23+x=-16

=>23+x=-16-16

=>23+x=-32

=>x=-32-23

=>x=-55

2,2x-35=15

=>2x=15+35

=>2x=50

=>x=25

3,3x+17=12

=>3x=12-17

=>3x=-5

=>ko có x thỏa mãn

4,|x-1|=0

=>x-1=0

=>x=1

5,-13.|x|=-26

=>|x|=-26:-13

=>|x|=2

=>x=2 hoặc -2

1. -16+23+x=-16

23+x=(-16)-(-16)

23+x=(-16)+16

23+x=0

x=0-23

x=-23

2. 2x-35=15

2x=15+35

2x=50

x=50:2

x=25

3. 3x+17=12

3x=12-17

3x=-5

x=(-5):3

x=-5/3

4. /x-1/=0

/x/=0+1

/x/=1

=>x=1 hoặc x=-1

5. -13./x/=-26

/x/=(-26):(-13)

/x/=2

=>x=2 hoặc x=-2

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

-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 + 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 ...