a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)

<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)

<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)

<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)

<=>\(4x-17=0\)

<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)

b) =>\(\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\3-\dfrac{1}{2}x=0\\1+\dfrac{3}{4}:x=0\end{matrix}\right.=>\left\{{}\begin{matrix}3x=\dfrac{1}{2}\\\dfrac{1}{2}x=3\\\dfrac{3}{4x}=-1\end{matrix}\right.=>\left\{{}\begin{matrix}x=\dfrac{1}{6}\\x=6\\x=\dfrac{-3}{4}\end{matrix}\right.\)

b: Đặt \(x^2-6x-2=a\)

Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)

=>(a+2)(a+7)=0

\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)

=>x(x-6)(x-1)(x-5)=0

hay \(x\in\left\{0;1;6;5\right\}\)

c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)

\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)

\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)

\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)

=>26x=-3

hay x=-3/26

a) Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)

\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)

\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)

\(\Leftrightarrow x+8-12+20x=0\)

\(\Leftrightarrow21x-4=0\)

\(\Leftrightarrow21x=4\)

\(\Leftrightarrow x=\dfrac{4}{21}\)

Vậy: \(S=\left\{\dfrac{4}{21}\right\}\)

Hình như em viết công thức bị lỗi rồi. Em cần chỉnh sửa lại để được hỗ trợ tốt hơn!

a) 

PT \(\Leftrightarrow \frac{4x+2}{12}-\frac{3x-6}{12}=\frac{12-8x}{12}-\frac{12x}{12}\)

\(\Leftrightarrow 4x+2-3x+6=12-8x-12x\)

\(\Leftrightarrow 21x=4\Leftrightarrow x=\frac{4}{21}\)

b) 

PT \(\Leftrightarrow \frac{30x+15}{20}-\frac{100}{20}-\frac{6x+4}{20}=\frac{24x-12}{20}\)

\(\Leftrightarrow 30x+15-100-6x-4=24x-12\Leftrightarrow -89=-12\) (vô lý)

Vậy pt vô nghiệm.

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

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

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

b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)

=>(7x+10)(x-3)=0

hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)

d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)

\(\Leftrightarrow x^2+14x+68=0\)

hay \(x\in\varnothing\)

\(a,3-x=x+1,8\)

\(\Rightarrow-x-x=1,8-3\)

\(\Rightarrow-2x=-1,2\)

\(\Rightarrow x=0,6\)

\(b,2x-5=7x+35\)

\(\Rightarrow2x-7x=35+5\)

\(\Rightarrow-5x=40\)

\(\Rightarrow x=-8\)

\(c,2\left(x+10\right)=3\left(x-6\right)\)

\(\Rightarrow2x+20=3x-18\)

\(\Rightarrow2x-3x=-18-20\)

\(\Rightarrow-x=-38\)

\(\Rightarrow x=38\)

\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)

\(\Rightarrow8x-3+1=1+6x+x\)

\(\Rightarrow8x-3=7x\)

\(\Rightarrow8x-7x=3\)

\(\Rightarrow x=3\)

\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)

\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)

\(\Rightarrow-2x=\dfrac{10}{9}\)

\(\Rightarrow x=-\dfrac{5}{9}\)

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

\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)

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

\(\Rightarrow x=\dfrac{16}{3}\)

\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)

\(\Rightarrow x-4=5-x\)

\(\Rightarrow x+x=5+4\)

\(\Rightarrow2x=9\)

\(\Rightarrow x=\dfrac{9}{2}\)

\(k,7x^2-11=6x^2-2\)

\(\Rightarrow7x^2-6x^2=-2+11\)

\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

\(m,5\left(x+3\cdot2^3\right)=10^2\)

\(\Rightarrow5\left(x+24\right)=100\)

\(\Rightarrow x+24=20\)

\(\Rightarrow x=-4\)

\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)

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

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)

\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)

#\(Urushi\text{☕}\)

a: 3-x=x+1,8

=>-2x=-1,2

=>x=0,6

b: 2x-5=7x+35

=>-5x=40

=>x=-8

c: 2(x+10)=3(x-6)

=>3x-18=2x+20

=>x=38

d; 8(x-3/8)+1=6(1/6+x)+x

=>8x-3+1=1+6x+x

=>8x-2=7x+1

=>x=3

e: =>-3x+x=4/3-2/9

=>-2x=12/9-2/9=10/9

=>x=-5/9

g: =>3/4x-1/2x=5/6+1/2

=>1/4x=5/6+3/6=8/6=4/3

=>x=4/3*4=16/3

h: =>x-4=-x+5

=>2x=9

=>x=9/2