a: =>11(x-3)=6(x-5)

=>11x-33=6x-30

=>5x=3

=>x=3/5

b: =>(4/3-1/4x-5/12)-2x=8/5*5/3=8/3

=>-9/4x+11/12=8/3

=>-9/4x=32/12-11/12=21/12=7/4

=>x=-7/9

c: =>1/2x-1/3-2/3x-1=x

=>-1/6x-4/3=x

=>-7/6x=4/3

=>x=-4/3:7/6=-4/3*6/7=-24/21=-8/7

d: =>1-2x-3x+1=7/2

=>-5x=3/2

=>x=-3/10

Câu 1: 

=>15(2x+1)-8(3x-1)=100

=>30x+15-24x+8=100

=>6x+23=100

hay x=77/6

Câu 2:

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

=>10x-6+12-21x+3-x-2=0

=>-12x=-7

hay x=7/12

Câu 3: 

\(\Leftrightarrow2\left(x^2-1\right)+3\left(x+1\right)=2\left(x^2-4x+4\right)\)

\(\Leftrightarrow2x^2-2+3x+3-2x^2+8x-8=0\)

=>11x-7=0

hay x=-7/11

Câu 4:

(x - 4)^3/6 + 1 = x(x + 1)/2 - (x - 5)(x + 5)/3

<=> (x - 4)^3 + 6/6 = x^2 + x/2 - x^2 - 25/3

<=> (x - 4)^3 + 6/6 = 3x^2 + 3x - 2x^2 + 50/6

<=> (x - 4)^3 + 6 = 3x^2 + 3x - 2x^2 + 50

<=> x^3 - 12x^2 + 48x - 58 = x^2 + 3x + 50

<=> x^3 -13x^2 + 45x - 108 = 0

Đến đây bạn bấm máy nhẩm nghiệm là ra nhé

Câu 5:

3(x + 2)^3/5 - (x - 1)^2/10 = (x - 3)(x + 3)/2

<=> 6(x + 2)^3 - (x - 1)^2/10 = 5(x^2 - 9)/10

<=> 6(x + 2)^3 - (x - 1)^2 = 5(x^2 - 9)

<=> 6x^3 + 36x^2 + 72x + 48 - x^2 + 2x - 1 - 5x^2 + 45 = 0

<=> 6x^3 + 30x^2 + 74x + 92 = 0

Đến đây bạn bấm máy nhẩm nghiệm như câu 4 nhé

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

Th1 : \(x-\dfrac{1}{2}=0\)

         \(x=0+\dfrac{1}{2}\)

         \(x=\dfrac{1}{2}\)

Th2 : \(-3-\dfrac{x}{2}=0\)

         \(\dfrac{x}{2}=-3\)

         \(x=\left(-3\right)\cdot2\)

         \(x=-6\)

Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)

b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)

    \(x=\dfrac{5}{8}+\dfrac{1}{8}\)

   \(x=\dfrac{3}{4}\)

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

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

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

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

                      \(x=0\)

d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)

    \(x+\dfrac{1}{3}=-4\)

    \(x=-4-\dfrac{1}{3}\)

    \(x=-\dfrac{13}{3}\)

\(b,\Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}-\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7x}{10}\\ \Rightarrow\dfrac{1}{2}x-\dfrac{3}{5}x+\dfrac{7}{10}x=\dfrac{6}{5}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{6}{5}\Rightarrow x=2\\ c,\Rightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=-\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{7}{6}\\ \Rightarrow\dfrac{4x-6-5+3x}{6}=\dfrac{7}{6}\\ \Rightarrow7x-11=7\Rightarrow x=\dfrac{18}{7}\\ d,\Rightarrow\dfrac{2}{3x}+\dfrac{7}{x}=\dfrac{4}{5}+2+\dfrac{3}{12}=\dfrac{61}{20}\\ \Rightarrow\dfrac{23}{3x}=\dfrac{61}{20}\\ \Rightarrow183x=460\\ \Rightarrow x=\dfrac{460}{183}\\ e,\Rightarrow2\left(x-1\right)-\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left(2-x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

e: Ta có: \(\left(x-1\right)^2=2\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

`e)3/(3x)-3/12=4/5-(7/x-2)`

`<=>1/x-1/4=4/5-7/x+2`

`<=>8/x=1/4+4/5+2=61/20`

`<=>1/x=61/160`

`<=>x=160/61`

`f)1/(x-1)+(-2)/3(3/4-6/5)=5/(2-2x)`

`<=>1/(x-1)+5/(2x-2)=2/3(3/4-6/5)=-3/10`

`<=>7/(2x-1)=-3/10`

`<=>2x-1=-70/3`

`<=>2x=-67/3`

`<=>x=-67/6`

e) Ta có: \(\dfrac{3}{3x}-\dfrac{3}{12}=\dfrac{4}{5}-\left(\dfrac{7}{x}-2\right)\)

\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{4}-\dfrac{4}{5}+\dfrac{7}{x}-2=0\)

\(\Leftrightarrow\dfrac{8}{x}=\dfrac{61}{20}\)

hay \(x=\dfrac{160}{61}\)

f) Ta có: \(\dfrac{1}{x-1}+\dfrac{-2}{3}\left(\dfrac{3}{4}-\dfrac{6}{5}\right)=\dfrac{5}{2-2x}\)

\(\Leftrightarrow\dfrac{2}{2x-2}-\dfrac{1}{2}-\dfrac{4}{5}+\dfrac{5}{2x-2}=0\)

\(\Leftrightarrow\dfrac{7}{2x-2}=\dfrac{13}{10}\)

\(\Leftrightarrow2x-2=\dfrac{70}{13}\)

\(\Leftrightarrow2x=\dfrac{96}{13}\)

hay \(x=\dfrac{48}{13}\)

a) Ta có: \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)

\(\Leftrightarrow\dfrac{3\left(2x-1\right)}{15}-\dfrac{5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)

\(\Leftrightarrow6x-3-5x+10-x-7=0\)

\(\Leftrightarrow0x=0\)(luôn đúng)

Vậy: S={x|\(x\in R\)}

Ai giúp vs

\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)

\(\Rightarrow2x=\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{1}{10}\)

\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=2\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)

\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)

\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)

\(\Leftrightarrow x=-\dfrac{49}{8}\)

\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)

\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)

\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)

\(\Leftrightarrow x=\dfrac{413}{160}\)

a= 1/10

b= 18/5

c= -49/8

d= 413

.

Giải:

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

     \(\left(\dfrac{7}{2}+2x\right).\dfrac{11}{3}=\dfrac{16}{3}\) 

                 \(\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{11}{3}\) 

                 \(\dfrac{7}{2}+2x=\dfrac{16}{11}\) 

                         \(2x=\dfrac{16}{11}-\dfrac{7}{2}\) 

                         \(2x=\dfrac{-45}{22}\) 

                           \(x=\dfrac{-45}{22}:2\) 

                           \(x=\dfrac{-45}{44}\) 

b) \(3-\left(17-x\right)=-12\) 

       \(3-17+x=-12\) 

                     \(x=-12-3+17\) 

                     \(x=2\) 

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

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

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

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

              \(x=\dfrac{-3}{5}\) 

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

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

             \(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{4}\) 

                 \(\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{4}:2\) 

                 \(\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{8}\) 

Vì giá trị tuyệt đối của 1 số nguyên ko bao giờ là số âm nên \(x\in\varnothing\) 

e) \(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}-\left(-1\right)=\dfrac{1}{3}\) 

                \(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}=\dfrac{1}{3}+\left(-1\right)\) 

                \(\left(-0,6x-\dfrac{1}{2}\right).\dfrac{3}{4}=\dfrac{-2}{3}\) 

                           \(-0,6x-\dfrac{1}{2}=\dfrac{-2}{3}:\dfrac{3}{4}\) 

                           \(-0,6x-\dfrac{1}{2}=\dfrac{-8}{9}\) 

                                   \(-0,6x=\dfrac{-8}{9}+\dfrac{1}{2}\) 

                                   \(-0,6x=\dfrac{-7}{18}\) 

                                           \(x=\dfrac{-7}{18}:-0.6\) 

                                           \(x=\dfrac{35}{54}\) 

f) \(\left(3x-1\right).\left(\dfrac{-1}{2}x+5\right)=0\) 

\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\\dfrac{-1}{2}x+5=0\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\) 

g) \(60\%.x+\dfrac{2}{3}=\dfrac{1}{3}.6\dfrac{1}{3}\) 

        \(\dfrac{3}{5}.x+\dfrac{2}{3}=\dfrac{1}{3}.\dfrac{19}{3}\) 

        \(\dfrac{3}{5}.x+\dfrac{2}{3}=\dfrac{19}{9}\) 

               \(\dfrac{3}{5}.x=\dfrac{19}{9}-\dfrac{2}{3}\) 

               \(\dfrac{3}{5}.x=\dfrac{13}{9}\) 

                    \(x=\dfrac{13}{9}:\dfrac{3}{5}\) 

                   \(x=\dfrac{65}{27}\) 

Chúc bạn học tốt!

f)câu khó nhất

=>3x-1=0 và -1/2x+5=0

   =>x=1/3 và x=10

\(b.3-\left(17-x\right)=-12\\ \Leftrightarrow17-x=15\\ \Rightarrow x=2\)

\(c.\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{1}{10}\\ \Leftrightarrow\dfrac{2}{3}x=-\dfrac{2}{5}\\ \Rightarrow x=-\dfrac{3}{5}\)

4, \(\Leftrightarrow4x+4+9\left(2x+1\right)=4x+6\left(x+1\right)+7+12x\)

\(\Leftrightarrow22x+13=22x+13\)vậy pt có vô số nghiệm 

5, \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=4-\dfrac{x}{3}\Rightarrow4x+2x-1=24-2x\)

\(\Leftrightarrow8x=25\Leftrightarrow x=\dfrac{25}{8}\)

6, \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\Rightarrow6x-6+3x-3=12-8\left(x-1\right)\)

\(\Leftrightarrow9x-9=20-8x\Leftrightarrow17x=29\Leftrightarrow x=\dfrac{29}{17}\)

\(a.x+\dfrac{1}{6}=-\dfrac{3}{8}\)

\(\Leftrightarrow x=-\dfrac{13}{24}\)

\(b.2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)

\(\Leftrightarrow2-\dfrac{3}{4}+x=\dfrac{7}{12}\)

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

\(c.\dfrac{1}{2}x+\dfrac{1}{8}x=\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{5}{8}x=\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{6}{5}\)

\(d.75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}-\left(\dfrac{-1}{2}\right)^2\)

\(=\dfrac{75}{100}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}-\dfrac{1}{4}\)

\(=-\dfrac{3}{4}+\dfrac{6}{5}-\dfrac{1}{4}\)

\(=\dfrac{1}{5}\)

a) \(x+\dfrac{1}{6}=\dfrac{-3}{8}\)

            \(x=\dfrac{-3}{8}-\dfrac{1}{6}\)

           \(x=\dfrac{-13}{24}\)

vậy x =....

b) \(2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)

             \(\dfrac{3}{4}-x=2-\dfrac{7}{12}\)

             \(\dfrac{3}{4}-x=\dfrac{17}{12}\)

                    \(x=\dfrac{3}{4}-\dfrac{17}{12}\)

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

vậy x =....