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

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

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

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

`2)x^4+2x^3-x^2-2x+1=0`

`<=>x^4+2x^3+x^2-2x^2-2x+1=0`

`<=>(x^2+x)^2-2(x^2+x)+1=0`

`<=>(x^2+x-1)^2=0`

`<=>x^2+x-1=0`

`\Delta=1+4=5`

`=>x_{1,2}=(-1+-sqrt5)/2`

Vậy `S={(-1+sqrt5)/2,(-1+sqrt5)/2`

`3)x^4-4x^3-9x^2+8x+4=0`

`<=>x^4-x^3-3x^3+3x^2-12x^2+12x-4x+4=0`

`<=>(x-1)(x^3-3x^2-12x-4)=0`

`<=>(x-1)(x^3+2x^2-5x^2-10x-2x-4)=0`

`<=>(x-1)(x+2)(x^2-5x-10)=0`

`+)x=1`

`+)x=-2`

`+)x^2-5x-10=0`

`Delta=25+40=65`

`=>x_{12}=(5+sqrt{65})/2`

\(\left(x-\frac{1}{5}\right)^2=\frac{1}{25}\)

\(\Leftrightarrow x-\frac{1}{5}=\frac{1}{5}\)

\(\Rightarrow x=\frac{2}{5}\)

P/s tham khảo nha

\(\left(\frac{3}{2}-x\right)^3=-8\)

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

\(\Rightarrow x=\frac{7}{2}\)

P/s tham khảo nha

\(\left(x-\frac{1}{5}\right)^2=\left(\frac{1}{5}\right)^2=\left(-\frac{1}{5}\right)^2\)

\(\orbr{\begin{cases}x-\frac{1}{5}=\frac{1}{5}\\x-\frac{1}{5}=-\frac{1}{5}\end{cases}}\)

\(\orbr{\begin{cases}x=\frac{2}{5}\\x=0\end{cases}}\)

Vậy...

b)\(\left(\frac{3}{2}-x\right)^3=-2^3\)

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

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

\(x=\frac{7}{2}\)

a) 3x-2/3 - 2 = 4x+1/4

<=>3x-8/3=4x+1/4

<=>3x-8/3-4x-1/4=0

<=>-x-29/12=0

<=>-x=29/12

<=>x=-29/12

Vậy x=-29/12

b) x-3/4 + 2x-1/3 = 2-x/6

<=>3x-13/12=2-x/6

<=>3x-13/12-2+x.1/6=0

<=> 19/6x-37/12=0

<=>19/6x=37/12

<=>x=37/38

Vậy x=37/38

x^3-3x^2+5x+2007=0

nên \(x\simeq-11,57\)

y^3-3y^2+5y-2013=0

nên \(y\simeq13,57\)

=>x+y=2

 1)    \(25x^4-10x^2y+y^2\)

\(\Leftrightarrow\left(5x^2\right)^2+2\cdot\left(5x^2\right)\cdot y+y^2\)

\(\Leftrightarrow\left(5x^2+y\right)^2\)

 2)   \(x^4+2x^3-4x-4\)

\(\Leftrightarrow\left(x^4-4\right)+\left(2x^3-4x\right)\Leftrightarrow\left(x^2-2\right)\left(x^2+2\right)+2x\left(x^2-2\right)\)

 \(\Leftrightarrow\left(x^2-2\right)\left(x^2+2+2x\right)\)

 3)  \(x^4+x^2+1\)

\(\Leftrightarrow x^4+x^2-x+x+1\)

 \(\Leftrightarrow\left(x^4-x\right)+\left(x^2+x+1\right)\)

\(\Leftrightarrow x\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2-x+1\right)\)

 4)    \(x^3-5x^2-14x\)\(\Leftrightarrow x^3-7x^2+2x^2-14x\)

\(\Leftrightarrow x^2\left(x-7\right)+2x\left(x-7\right)\)\(\Leftrightarrow x\left(x+2\right)\left(x-7\right)\)

 5)  \(x^2yz+5xyz-14yz\)\(\Leftrightarrow yz\left(x^2+5x-14\right)\)

\(\Leftrightarrow yz\left(x^2+7x-2x-14\right)\)

\(\Leftrightarrow yz\left[x\left(x+7\right)-2\left(x+7\right)\right]\) 

\(\Leftrightarrow yz\left(x+7\right)\left(x-2\right)\)

Cảm ơn bạn Nguyễn Kim Thương :))