\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)

\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)

\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2.\\x=1.\end{matrix}\right.\)

\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)

\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)

\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)

\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)

\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)

\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)

\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

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

1) \(\left(x-3\right)^2-4=0\)

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

\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

2) \(x^2-2x=24\)

\(\Leftrightarrow x^2-2x-24=0\)

\(\Leftrightarrow x^2+4x-6x-24=0\)

\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

Câu 3 số xấu rồi e

alo =0 nhé anh pop pop

\(\left(x-\dfrac{3}{2}\right)\times\left(2x+1\right)>0\)

Th1:

\(x-\dfrac{3}{2}>0\Leftrightarrow x>\dfrac{3}{2}\)

\(2x+1>0\Leftrightarrow2x>1\Leftrightarrow x>\dfrac{1}{2}\)

( 1 )

Th2: 

\(x-\dfrac{3}{2}< 0\Leftrightarrow x< \dfrac{3}{2}\)

\(2x+1< 0\Leftrightarrow2x< -1\Leftrightarrow x< -\dfrac{1}{2}\)

( 2 )

Từ ( 1 ) và ( 2 ), ta có:

\(\Rightarrow x< -\dfrac{1}{2};x>\dfrac{3}{2}\)

\(\left(2-x\right)\times\left(\dfrac{4}{5}-x\right)< 0\)

Th1:

\(2-x>0\Leftrightarrow x>2\)

\(\dfrac{4}{5}-x< 0\Leftrightarrow x< \dfrac{4}{5}\)

( Loại )

Th2:

\(2-x< 0\Leftrightarrow x< 2\)

\(\dfrac{4}{5}-x>0\Leftrightarrow x>\dfrac{4}{5}\)

=> \(\dfrac{4}{5}< x< 2\)

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`<=>(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`

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

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`+)x=1`

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`Delta=25+40=65`

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