a: =>x^2-25-x^2-3x=10

=>-3x=35

=>x=-35/3

b: =>4x^2-9-4(x^2+4x+4)=5

=>4x^2-9-4x^2-16x-16-5=0

=>-16x-30=0

=>x=-15/8

c: =>9x^2+45x-9x^2+4=7

=>45x=3

=>x=1/15

d: =>x^3+3x^2+3x+1-x^3-3x^2+5x=8

=>8x=7

=>x=7/8

a)

   \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)

   \(\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}\)

    \(\frac{1}{3}:x=\frac{-1}{15}\)

     \(x=\frac{1}{3}:\frac{-1}{15}\)

    \(x=-5\)

b)

     \(\frac{10}{3x}+\frac{67}{4}=-13.25\)

      \(\frac{10}{3x}+\frac{67}{4}=\frac{-53}{4}\)

       \(\frac{10}{3x}=\frac{-53}{4}+\frac{67}{4}\)

       \(\frac{10}{3x}=\frac{7}{2}\)

      \(\Rightarrow10.2=7.3x\)

            \(20=21x\)

            \(x=\frac{20}{21}\)

c)

        x+30%x=-1.3

        x+0.3x=-1.3

        x (1+0.3) = -1.3

        x . 1.3 = -1.3

        x = -1.3 : 1.3

        x = 1

d)

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

           \(\frac{14}{5x}-50=51.\frac{2}{3}\)

             \(\frac{14}{5x}-50=34\)

             \(\frac{14}{5x}=34+50\)

              \(\frac{14}{5x}=84\)

               \(84.5x=14\)

              420x = 14

                   \(x=\frac{1}{30}\)

a, \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)

=> \(\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}\)

=>\(\frac{1}{3}:x=\frac{-1}{15}\)

=> \(x=\frac{1}{3}:\frac{-1}{15}\)

=> \(x=\frac{-1}{5}\)

mình sai chỗ kết quả cuối bằng -5 nhé

\(a,\frac{-3}{2}-2x+\frac{3}{4}=-1\)

\(\frac{-3}{2}-2x=-1-\frac{3}{4}\)

\(\frac{-3}{2}-2x=\frac{-7}{4}\)

\(2x=\frac{-7}{4}+\frac{-3}{2}\)

\(2x=\frac{-13}{4}\)

\(x=\frac{-13}{4}:2\)

\(x=\frac{-13}{4}.\frac{1}{2}\)

\(x=\frac{-13}{8}\)

(−23×x−35)×(3−2−103)=25(−23×x−35)×(3−2−103)=25

⇔(−23×x−35)×−296=25⇔(−23×x−35)×−296=25

⇔−23×x−35=25÷−296⇔−23×x−35=25÷−296

⇔−23×x−35=−12145⇔−23×x−35=−12145 

⇔−23×x=−12145+35⇔−23×x=−12145+35

⇔−23×x=1529⇔−23×x=1529 

⇔x=1529÷−23⇔x=1529÷−23

⇔x=−4558⇔x=−4558 

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

\(x\left(\dfrac{2}{3}-\dfrac{3}{2}\right)=\dfrac{5}{12}\)

\(x\cdot\left(-\dfrac{5}{6}\right)=\dfrac{5}{12}\)

\(x=\dfrac{5}{12}:\left(-\dfrac{5}{6}\right)\)

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

Vậy \(x=-\dfrac{1}{2}\).

b) \(\dfrac{2}{5}+\dfrac{3}{5}\cdot\left(3x-3\cdot7\right)=-\dfrac{53}{10}\)

\(\dfrac{3}{5}\left(3x-3\cdot7\right)=-\dfrac{53}{10}-\dfrac{2}{5}\)

\(\dfrac{3}{5}\left(3x-3\cdot7\right)=-\dfrac{57}{10}\)

\(3x-3\cdot7=-\dfrac{57}{10}:\dfrac{3}{5}\)

\(3x-3\cdot7=-\dfrac{19}{2}\)

\(3x-21=-\dfrac{19}{2}\)

\(3x=-\dfrac{19}{2}+21\)

\(3x=\dfrac{23}{2}\)

\(x=\)\(\dfrac{23}{2}:3\)

\(x=\dfrac{23}{6}\)

Vậy \(x=\dfrac{23}{6}\).

c) \(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)+\dfrac{5}{3}=\dfrac{23}{27}\)

\(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)=\dfrac{23}{27}-\dfrac{5}{3}\)

\(\dfrac{7}{9}:\left(2+\dfrac{3}{4x}\right)=-\dfrac{22}{27}\)

\(2+\dfrac{3}{4x}=\dfrac{7}{9}:-\dfrac{22}{27}\)

\(2+\dfrac{3}{4x}=-\dfrac{21}{22}\)

\(\dfrac{3}{4x}=-\dfrac{21}{22}-2\)

\(\dfrac{3}{4x}=-\dfrac{65}{22}\)

\(4x=\dfrac{3\cdot22}{-65}\)

\(4x=-\dfrac{66}{65}\)

\(x=-\dfrac{66}{65}:4\)

\(x=-\dfrac{33}{130}\)

Vậy \(x=-\dfrac{33}{130}\).

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

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

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

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

\(x=-\dfrac{3}{20}\)

Vậy \(x=-\dfrac{3}{20}\).

e) \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)

\(\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}\)

\(\left|x\right|=\dfrac{29}{12}\)

\(x=\dfrac{29}{12}\) hoặc \(=-\dfrac{29}{12}\)

Vậy \(x\in\left\{\dfrac{29}{12};-\dfrac{29}{12}\right\}\).

         Bài 1:

- \(\dfrac{11}{2}x\) + 1 = \(\dfrac{1}{3}x-\dfrac{1}{4}\)

- \(\dfrac{11}{2}\)\(x\) - \(\dfrac{1}{3}\)\(x\) = - \(\dfrac{1}{4}\) - 1

-(\(\dfrac{33}{6}\) + \(\dfrac{2}{6}\))\(x\) = - \(\dfrac{5}{4}\)

- \(\dfrac{35}{6}\)\(x\) = - \(\dfrac{5}{4}\)

  \(x=-\dfrac{5}{4}\) : (- \(\dfrac{35}{6}\))

 \(x\) = \(\dfrac{3}{14}\)

Vậy \(x=\dfrac{3}{14}\)

Bài 2: 2\(x\) - \(\dfrac{2}{3}\) - 7\(x\) = \(\dfrac{3}{2}\) - 1

         2\(x\) - 7\(x\) = \(\dfrac{3}{2}\) - 1 + \(\dfrac{2}{3}\)

         - 5\(x\)    = \(\dfrac{9}{6}\) - \(\dfrac{6}{6}\) + \(\dfrac{4}{6}\) 

        - 5\(x\)    = \(\dfrac{7}{6}\)

           \(x\)    = \(\dfrac{7}{6}\) : (- 5) 

          \(x\)    = - \(\dfrac{7}{30}\)

Vậy \(x=-\dfrac{7}{30}\)

Bài 3: \(\dfrac{3}{2}\)\(x\) - \(\dfrac{2}{5}\) = \(\dfrac{1}{3}x\) - \(\dfrac{1}{4}\)

           \(\dfrac{3}{2}\)\(x\) - \(\dfrac{1}{3}x\) = \(-\dfrac{1}{4}\) + \(\dfrac{2}{5}\)

           (\(\dfrac{9}{6}\) - \(\dfrac{2}{6}\))\(x\) = \(\dfrac{-5}{20}\) + \(\dfrac{8}{20}\)

            \(\dfrac{7}{6}x\)       = \(\dfrac{3}{20}\)

            \(x\)       = \(\dfrac{3}{20}\) : \(\dfrac{7}{6}\)

            \(x\)      = \(\dfrac{9}{70}\)

Vậy  \(x=\dfrac{9}{70}\)

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