a; -2\(x\) - 3.(\(x-17\)) = 34 - 2.( - \(x\) + 25)

    - 2\(x\) - 3\(x\) + 51 = 34 + 2\(x\) - 50

       2\(x\) + 2\(x\) + 3\(x\) = - 34 + 50 + 51

            7\(x\)            = 67

               \(x\)           =  67 : 7

                \(x\)         =  \(\dfrac{67}{7}\)

Vậy \(x\) = \(\dfrac{67}{7}\) 

b; 17\(x\) + 3.(- 16\(x\) - 37) = 2\(x\) + 43 - 4\(x\)

    17\(x\) - 48\(x\)  - 111 =  2\(x\) - 4\(x\) + 43

    - 31\(x\) - 2\(x\) + 4\(x\) = 111 + 43

        - \(x\) x (31 + 2 - 4) = 154

        - \(x\) x (33 - 4) =  154

         - \(x\) x 29 = 154

         - \(x\)         = 154 : (-29)

           \(x\)        = - \(\dfrac{154}{29}\)

          Vậy \(x=-\dfrac{154}{29}\) 

c; {-3\(x\) + 2.[45 - \(x\) - 3(3\(x\) + 7) - 2\(x\)] + 4\(x\) } = 55

    {-3\(x\) + 2.[45 - \(x\) - 9\(x\) - 21 - 2\(x\)] + 4\(x\)} = 55

    {- 3\(x\) + 2.[(45 - 21) - (\(x+9x\)+2\(x\))] + 4\(x\) } = 55

     { (- 3\(x\) + 4\(x\)) + 2.[24 - 12\(x\)] } = 55

                   \(x\) + 48 - 24\(x\) = 55

                   \(x-24x\)         = 55 - 48

                     - 23\(x\)           = 7

                            \(x\)           = 7 : - 23

                             \(x=-\dfrac{7}{23}\)

Vậy \(x=-\dfrac{7}{23}\) 

\(a,-5.\left(-x+7\right)-3.\left(-x-5\right)=-4.\left(12-x\right)+48\)

                          \(5x-35-3x+15=-48+4x+48\)

                                                \(2x-10=4x\)

                                                \(2x-4x=10\)

                                                       \(-2x=10\)

                                                             \(x=-5\)

 \(b,\left(-x-7\right)-5.\left(-x-3\right)=12.\left(3-x\right)\)

                  \(-x-7+5x+15=36-12x\)

                                         \(4x+8=36-12x\)

                                    \(4x+12x=36-8\)

                                               \(16x=28\)

                                                    \(x=1,75\)

các câu còn lại tương tự nha

a ) \(-5\times\left(-x+7\right)-3\times\left(-x-5\right)=-4\times\left(12-x\right)+48\)

\(\Leftrightarrow5x-35+3x+15=-48+4x+48\)

\(\Leftrightarrow5x-3x+4x=35-15-48+48\)

\(\Leftrightarrow2x=20\)

\(\Leftrightarrow x=10\)

b ) \(-2\times\left(15-3x\right)-4\times\left(-7x+8\right)=-5-9\times\left(-2x+1\right)\)

\(\Leftrightarrow-30+6x-28x-32=-5+18x-9\)

\(\Leftrightarrow6x-28x-18x=30+32-5-9\)

\(\Leftrightarrow-40x=48\)

\(\Leftrightarrow x=-1.2\)

C,D nưã bạn

 c, 7 . ( - x - 7 ) - 5 . ( - x - 3 ) = 12 . ( 3 - x)

<=> - 7x - 49  + 5x  + 15 = 36 - 12x

<=>  - 2x + 12x = 36 - 15 + 49

<=> 10 x = 70

<=> x = 7

Vẫy  x = 7

d, 5 . ( - 3x - 7 ) - 4 . ( - 2x - 11 ) = 7 . ( 4x + 10) + 9

<=> - 15x + 8x + 44 = 28x + 70 + 9

<=> - 7x + 44 = 28x + 79

<=> - 7x - 28x = 79 - 44

<=> - 35x = 35

<=> x = - 1

Vậy x = - 1

Hc tốt

Hướng dẫn cách làm:

B1:Bạn phá ngoặc ra

B2:Bạn chuyển hết x sang một bên(Nhớ sử dụng quy tắc chuyển vế đổi dấu)

B3:Bạn tính hai vế

(Áp dụng quy tắc chuyển vế đổi dấu)

(Hoặc)

B1:Bạn tính hai vế

B2:Chuyển hết x sang một vế

B3:Tính vế chứa x

(Áp dụng quy tắc lấy tiềnmở khóa cổng-mở khóa cửa-mở khóa két-thò tay vào lấy tiền)

P/s:Khi tìm x nhớ áp dụng hai quy tắc mình nói trên

a) -5.(-x+7)-3.(-x-5)= -4.(12-x)+48
5x - 35 +3x +15 = -48 + 4x +48
5x + 3x - 4x = 35 - 15 -48 + 48
2x = 20
x = 10
b) (-x-7)-5.(-x-3)= 12.(3-x)
-x-7+5x+15 = 36 -12x
-x + 5x +12x = 7 -15 +36
16x = 28
x = 1.75
c) -2.(15-3x)-4.(7x+8) = -5-9.(-2x+1)
-30 + 6x -28x -32 = -5 +18x -9
6x -28x - 18x = 30+32-5-9
-40x = 48
x = -1.2
d) (-3x-7)-4.(-2x-11)= 7.(4x+10)+9
-3x -7 + 8x + 44 = 28x+70 +9
-3x +8x-28x = 7-44 +70+9
-23x =42
x = -42/23

a: 2x-3/2+3/4=-4

=>2x-3/4=-4

=>2x=-13/4

hay x=-13/8

b: \(\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)\cdot\left(\dfrac{-3}{2}-\dfrac{10}{3}\right)=\dfrac{2}{5}\)

\(\Leftrightarrow-\dfrac{2}{3}x-\dfrac{3}{5}=\dfrac{2}{5}:\dfrac{-29}{6}=\dfrac{-2}{5}\cdot\dfrac{6}{29}=\dfrac{-12}{145}\)

=>2/3x+3/5=12/145

=>2/3x=-15/29

hay x=-45/58

c: \(\dfrac{x}{2}-\left(\dfrac{3}{5}x-\dfrac{13}{5}\right)=-\left(\dfrac{7}{10}x+\dfrac{7}{5}\right)\)

=>1/2x-3/5x+13/5=-7/10x-7/5

=>-1/10x+7/10x=-7/5-13/5

=>3/5x=-2

hay x=-2:3/5=-10/3

a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)

\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)

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

b) \(\dfrac{39}{7}:x=13\)

\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)

c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)

\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)

\(\dfrac{14}{5}x=34+50=84\)

\(x=\dfrac{84}{\dfrac{14}{5}}=30\)

d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)

\(\dfrac{1}{6}x=\dfrac{5}{12}\)

\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)

g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)

\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)

\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)

\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)

\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)

h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)

\(x=1\)

Mỏi tay woa bn làm nốt nha!!

ai giúp minh đi cần quá gấp

1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm