\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)

\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)

\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)

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

hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)

1)

a) \(2x-6=0\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

b) \(x\times\left(x+2\right)-3\times\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\times\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

c) \(\frac{x-6}{x+1}=\frac{x^2}{x-1}\)

nhân chéo lên, ngại chết đc

\(x^5-x^3+x^2-1=x^3\left(x^2-1\right)+\left(x^2-1\right)=\left(x^2-1\right)\left(x^3+1\right)=\left(x-1\right)\left(x+1\right)^2\left(x^2-x+1\right)\)

Ta có 

1/6=4/24

1/7=4/28

=> 4/24>x>4/28

=>x =4/25 hoặc 4/26 hoặc 4/27

                 Đáp số 4/25;4/26;4/27

1/3 + 1/6 + \(\dfrac{x}{4042}=1\)

\(\Leftrightarrow\dfrac{x}{4042}=1-\dfrac{1}{3}-\dfrac{1}{6}\)

\(\dfrac{x}{4042}=\dfrac{6}{6}-\dfrac{2}{6}-\dfrac{1}{6}\)

\(\dfrac{x}{4042}=\dfrac{3}{6}\)

\(x=\dfrac{3}{6}\cdot4042\)

\(x=2021\)

a)Ta có:5/3.x^2-1/2.x^2y

          =(5/3-1/2).x^2y

          = 7/6.x^2y(Bậc 3)

b)Ta có: 7/6.(-2)^2(-1)

             = 7/6.4.(-1)

             = 7/6.(-4)

             =-28/6

a, -  A=\(\dfrac{5}{3}\).x².y-\(\dfrac{-1}{2}\).x².y

      =\(\dfrac{13}{6}\).x².y

- Bậc= 3.

b, A=\(\dfrac{13}{6}\).(-2)².(-1)

      =\(\dfrac{13}{6}\).4.(-1)

      =\(\dfrac{-26}{3}\)

bạn Tt_Cindy_tT làm sai rồi 5/3-1/2=7/6 chứ ko bằng 13/6

Ta có: \(3\left|x^2-1\right|-6=\left|1-x^2\right|\)

\(\Leftrightarrow3\left|x^2-1\right|-\left|x^2-1\right|=6\)

\(\Leftrightarrow2\left|x^2-1\right|=6\)

\(\Leftrightarrow\left|x^2-1\right|=3\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-1=3\\x^2-1=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x^2=4\\x^2=-2\end{cases}}\)

Vì \(x\ge0>-2\left(\forall x\right)\)

\(\Rightarrow x^2=4\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

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

`3xx(x-1/6)-2xx(x-1/3)=2/3`

`3x-1/2-2x+2/3=2/3`

`x=2/3-2/3+1/2`

`x=1/2`

`3x(x-1/6)-2x(x-1/3)=2/3`

`3x^2 - 1/2 x- 2x^2 + 2/3x = 2/3`

`x^2+1/6 x =2/3`

`x=(-1 \pm \sqrt97)/12`