C

a) /x+2/=0

=>x+2=0

=>x=0-2

=>x=-2

  \(\text{a) /x+2/=0}\)

\(\Rightarrow x=0-2\)

\(\Rightarrow x=2\)

\(\text{b) /x-5/=/-7/}\)

\(\Rightarrow\text{/x-5/=7}\)

\(\Rightarrow\orbr{\begin{cases}x-5=7\\x-5=-7\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=12\\x=-2\end{cases}}\)

\(\text{c) /x-3/=7-(-2)}\)

\(\Rightarrow/x-3/=9\)

\(\Rightarrow\orbr{\begin{cases}x=-9+3=-6\\x=9+3=12\end{cases}}\)

Bài làm

Câu a) Nguyễn Văn Cường Làm rồi nên tôi lm mấy câu còn lại

b) | x - 5 | = | - 7 |

=> | x - 5 | = 7

=> \(\hept{\begin{cases}x-5=-7\Rightarrow x=2\\x-5=7\Rightarrow x=12\end{cases}}\)

Vậy x = 2 hoặc x = 12

c) | x - 3 | = 7 - ( - 2 )

    | x - 3 | = 9

=> \(\hept{\begin{cases}x-3=9\Rightarrow x=12\\x-3=-9\Rightarrow x=-6\end{cases}}\)

Vậy x = 12 hoặc x = -6

d) ( 7 - x ) - ( 25 + 7 ) = -25

    ( 7 - x ) - 32           = - 25

      7 - x                    = -25 + 32

      7 - x                    = 7

           x                    = 7 - 7

           x                    = 0

Vậy x = 0

# Học tốt #

b) |x-3|=7 - (-2)

|x-3| = 9

TH 1 : TH 2 :

x - 3 = 9 x - 3 = -9

x = 9 + 3 x = -9 + 3

x = 12 x = 6

c) (7-x)-(25+7)=-25

(7-x) = -25 + 25 - 7

(7-x) = - 7

x = -7 + 7

x = 0

a) |x-5| = |-7|

|x-5| = 7

\(\Rightarrow\left[{}\begin{matrix}x-5=7\\x-5=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7+5\\x=\left(-7\right)+5\end{matrix}\right.\Rightarrow}\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

Vậy x \(\in\) { 12;-2}

b) |x-3| = 7 - ( - 2 )

|x-3| = 7 + 2

|x-3| = 9\(\Rightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9+3\\x=\left(-9\right)+3\end{matrix}\right.\Rightarrow}\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)

Vậy x \(\in\) { 12 ; -6 }

c) ( 7 - x ) - ( 25 + 7 ) = - 25

( 7 - x ) - 32 = - 25

7 - x = ( - 25 ) + 32

7 - x = 7

x = 7 - 7

x = 0

/x+2/=0\(\Rightarrow x+2=0\Rightarrow x=0-2=-2\)

/x-3/=7-(-2)=9\(\Rightarrow x=9+3=12ho\text{ặc}x=-9+3=-6\)

(7-x)-(25+7)=-25\(\Rightarrow\left(7-x\right)-32=-25\Rightarrow7-x=-25+32=7\Rightarrow x=0\)

/x-3/=/5/+/-7/=5+7=12\(\Rightarrow x=15ho\text{ặc}x=-9\)

/x-5/=/-7/=7\(\Rightarrow x=12ho\text{ặc}x=-2\)

4-(7-x)=x-(13-4)\(\Rightarrow x-3=x-9\Rightarrow x-x=-9+3\Rightarrow0=-6\)(vô lí)

Vậy không có x thoả mãn 4-(7-x)=x-(13-4)

|x+2|=0

Vì |0|=0,suy ra x+2=0

                         x=0-2

                         x=-2

Vậy x = -2

|x-3|=7-(-2)

|x-3|=7+2

|x-3|=9

Vì |9|=|-9|=9,suy ra x-3 thuộc{9;-9}

*x-3=9

x   =9+3

x   =12

*x-3=-9

    x=-9+3

     x=-6

Vậy x thuộc {12;-6}

(7-x)-(25+7)=-25

(7-x)-32      =-25

            7-x =-25+32

            7-x  =57

              x   =7-57

              x   =7+(-57)

              x    =-50

vậy x=-50

|x-3|=|5|+|-7|

|x-3|=5+7

|x-3|=12

Vì |12|=|-12|=12,suy ra x-3 thuộc {12;-12}

*x-3=12

  x=12+3

  x=15

*x-3=-12

    x=-12+3

    x=-9

vậy x thuộc {15;-9}

3/25 x ( 15/7 - 2/7 ) + 3/7 x 1/25 

= 3/25 x 13/7 + 3/7 x 1/25 

= (3 x 13/7 + 3/7 ) x 1/25 

= 42/7 x 1/25 

= 6 x 1/25 

= 6/25

\(\dfrac{3}{25}\times\dfrac{15}{7}+\dfrac{3}{7}\times\dfrac{1}{25}-\dfrac{2}{7}\times\dfrac{3}{25}\)

\(=\dfrac{3}{25}\times\left(\dfrac{15}{7}-\dfrac{2}{7}\right)+\dfrac{3}{7}\times\dfrac{1}{25}\)

\(=\dfrac{3}{25}\times\dfrac{13}{7}+\dfrac{3}{7}\times\dfrac{1}{25}\)

\(=\dfrac{3\times13}{25\times7}+\dfrac{3\times1}{7\times25}\)

\(=\dfrac{39}{175}+\dfrac{3}{175}\)

\(=\dfrac{39+3}{175}\)

\(=\dfrac{42}{175}\)

\(=\dfrac{6}{25}\)

A = \(\dfrac{3}{25}\) \(\times\) \(\dfrac{15}{7}\) + \(\dfrac{3}{25}\) \(\times\) \(\dfrac{1}{7}\) - \(\dfrac{2}{7}\times\) \(\dfrac{3}{25}\)

A = \(\dfrac{3}{25}\) \(\times\)( \(\dfrac{15}{7}\) + \(\dfrac{1}{7}-\dfrac{2}{7}\))

A = \(\dfrac{3}{25}\) \(\times\) \(\dfrac{14}{7}\)

A = \(\dfrac{6}{25}\)

1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)

\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)

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

\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)

2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)

\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)

\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)

3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)

\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)

\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)

\(=\dfrac{17}{7}\)

4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)

\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)

\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)

\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)

\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)