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1) -5+(-3+12-x)=7-(2-8)

=>-5+(9-x)=7-(-6)

=>-5+9-x=13

=>x=9-5-13=-9

2) 8-(7-x+3)=-|-7+2|-(-9)

=>8-(10-x)=-|-5|+9

=>8-10+x=-5+9

=>x=-5+9-8+10=6

3)-(12-5+9-x)=3-|5-8+2|

=>-(16-x)=3-|-1|

=>-16+x=3-1

=>x=3-1+16=18

4)-(-3-x+8)-(-3+5)=|5-8|

=>-(5-x)-2=|-3|

=>-5+x-2=3

=>x=3+5+2=10

5)-|5-8|-(4-x+5)=4+|3-9|

=>-|-3|-(9-x)=4+|-6|

=>-3-9+x=4+6

=>x=4+6+3+9=22

câu hỏi tương tự

Bài 1:

a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)

b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)

c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)

d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)

hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)

e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)

hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)

dad

Bài 1:

a) \(\dfrac{9}{20}-\dfrac{8}{15}\times\dfrac{5}{12}\)

\(=\dfrac{9}{20}-\dfrac{2}{9}\)

\(=\dfrac{41}{180}\)

b) \(\dfrac{2}{3}\div\dfrac{4}{5}\div\dfrac{7}{12}\)

\(=\dfrac{2}{3}\times\dfrac{5}{4}\times\dfrac{12}{7}\)

\(=\dfrac{5}{6}\times\dfrac{12}{7}\)

\(=\dfrac{10}{7}\)

c) \(\dfrac{7}{9}\times\dfrac{1}{3}+\dfrac{7}{9}\times\dfrac{2}{3}\)

\(=\dfrac{7}{9}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)

\(=\dfrac{7}{9}\times1\)

\(=\dfrac{7}{9}\)

Bài 2:

a) \(2\times\left(x-1\right)=4026\)

\(\left(x-1\right)=4026\div2\)

\(x-1=2013\)

\(x=2014\)

Vậy: \(x=2014\)

b) \(x\times3,7+6,3\times x=320\)

\(x\times\left(3,7+6,3\right)=320\)

\(x\times10=320\)

\(x=320\div10\)

\(x=32\)

Vậy: \(x=32\)

c) \(0,25\times3< 3< 1,02\)

\(\Leftrightarrow0,75< 3< 1,02\) ( S )

=> \(0,75< 1,02< 3\)

1: =>x=8-35=-27

2: =>15-4+x=6

=>x+11=6

hay x=-5

3: =>-30+25-x=-1

=>x+5=1

hay x=-4

4: =>x-(-13)=-8

=>x+13=-8

hay x=-21

5: =>x-29-17+38=-9

=>x-8=-9

hay x=-1

Giải:

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

     \(\dfrac{-5}{6}-x=\dfrac{1}{4}\)

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

               \(x=\dfrac{-13}{12}\) 

b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\) 

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

    \(2.\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)  

             \(x-\dfrac{1}{3}=\dfrac{2}{3}:2\) 

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

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

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

c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\) 

           \(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\) 

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

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\) 

d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\) 

\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\) 

            \(x.\dfrac{5}{6}=\dfrac{29}{8}\) 

                \(x=\dfrac{29}{8}:\dfrac{5}{6}\) 

                \(x=\dfrac{87}{20}\)

a) 3/8 . x = 9/8 - 1

3/8 . x = 1/8

x = 1/8 : 3/8

x = 1/3

b) 4/5 . x = 7/5 - 1/5

4/5 . x = 6/5

x = 6/5 : 4/5

x = 3/2

c) 12/7 : x + 2/3 = 7/5

12/7 : x = 7/5 - 2/3

12/7 : x = 11/15

x = 12/7 : 11/15

x = 180/77

d) 3.(x + 7) - 15 = 27

3.(x + 7) = 27 + 15

3.(x + 7) = 42

x + 7 = 42 : 3

x + 7 = 14

x = 14 - 7

x = 7

a) \(\dfrac{3}{8}x=\dfrac{9}{8}-1\)

\(\Rightarrow\dfrac{3}{8}x=\dfrac{1}{8}\)

\(\Rightarrow x=\dfrac{1}{8}:\dfrac{3}{8}=\dfrac{1}{3}\)

b) \(\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{1}{5}\)

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

\(\Rightarrow x=\dfrac{6}{5}:\dfrac{4}{5}=\dfrac{3}{2}\)

c) \(\dfrac{12}{7}:x+\dfrac{2}{3}=\dfrac{7}{5}\)

\(\Rightarrow\dfrac{12}{7}:x=\dfrac{7}{5}-\dfrac{2}{3}\)

\(\Rightarrow\dfrac{12}{7}:x=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{12}{7}:\dfrac{11}{15}=\dfrac{180}{77}\)

d) \(3\left(x+7\right)-15=27\)

\(\Leftrightarrow3\left(x+7\right)=42\)

\(\Leftrightarrow x+7=14\Leftrightarrow x=7\)

1) |x + 2| = 4

\(\Leftrightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)

2) 3 – |2x + 1| = (-5)

\(\Leftrightarrow\left|2x+1\right|=8\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}\)

3) 12 + |3 – x| = 9

\(\Leftrightarrow\left|3-x\right|=-3\)(vô lí)

=>\(x=\varnothing\) 

1) I x+2 I=4

\(\Rightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-6\end{cases}}}\)

2) \(3-|2x+1|=-5\)

\(\Leftrightarrow|2x+1|=8\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}}\)

3) \(12+|3-x|=9\)

\(\Leftrightarrow|3-x|=-3\)(vô lí vì I 3-x I \(\ge\)0)

1; |\(x\) + 2| = 4

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

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

     \(\left[{}\begin{matrix}x=-6\\x=2\end{matrix}\right.\)

Vậy \(x\) \(\in\) {- 6; 2}

1 , 10

2 , 8

3 , 5

4 , 2

5 , 1

6 ,

1/ `|x|=10<=> x=\pm 10`

2/ `|x-8|=0<=>x-8=0<=>x=8`

3/ `7+|x|=12<=>|x|=5<=>x=\pm 5`

4/ `|x+1|=3`

$\Leftrightarrow\left[\begin{array}{1}x+1=3\\x+1=-3\end{array}\right.\\\Leftrightarrow\left[\begin{array}{1}x=3\\x=-4\end{array}\right.$

5/ `15-x=16-(14-42)`

`<=>15-x=16+28`

`<=>15-x=44`

`<=>x=-29`

6/ `210-(x-12)=168`

`<=>210-x+12=168`

`<=>222-x=168`

`<=>x=54`

1.

\(\left|x\right|=10\Leftrightarrow x=\pm10\)

2.

\(\left|x-8\right|=0\Leftrightarrow x-8=0\Leftrightarrow x=8\)

3.

\(7+\left|x\right|=12\Leftrightarrow\left|x\right|=5\Leftrightarrow x=\pm5\)