sửa đề : 

\(x^3-3x^2+3x-1=0\)

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

      `x - ( 2x - 1 ) <= 3x - 3`

`<=> x - 2x + 1 <= 3x - 3`

`<=> 3x - x + 2x >= 1 + 3`

`<=> 4x >= 4`

`<=> x >= 1`

Vậy `S = { x | x >= 1 }`

\(\Leftrightarrow x-2x+1\le3x-3\)

\(\Leftrightarrow-4x\le-4\)

\(\Leftrightarrow x\ge1\)

`x - (2x - 1) <= 3x - 3`

`<=> x - 2x + 1 <= 3x - 3`

`<=> 3x - x + 2x >= 1 + 3`

`<=> 4x >= 4`

`<=>` `x >= 1`

Vậy `S = {x | x >= 1}`

`x(x - 4) - 3x + 12 = 0`

`<=> x(x - 4) + 3(x - 4) = 0`

`<=> (x + 3)(x - 4) = 0`

`<=>` $\left[\begin{matrix} x + 3 = 0\\ x - 4 = 0\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x = -3\\ x = 4\end{matrix}\right.$

Vậy `S = {-3; 4}`

\(x\left(x-4\right)-3x+12=0\)

\(\Leftrightarrow x^2-4x-3x+12=0\)

\(\Leftrightarrow x\left(x-4\right)-3\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

\(x+x^2=0\)

\(\Leftrightarrow x\left(1+x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\1+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Vậy: Phương trình có tập nghiệm \(S=\left\{0;-1\right\}\)

\(x\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Tham khảo thử đúng không nha mn

     \(x^2+x-y^2=0\)

⇔ \(\left(x^2-y^2\right)+x=0\)

⇔ \(\left(x-y\right)\left(x+y\right)+x=0\)

⇒ \(x-y=0\) hoặc \(x+y=0\) hoặc \(x=0\)

⇒ \(x=y=0\)

\(\Leftrightarrow x^2+x=y^2\)

\(\Leftrightarrow4x^2+4x=4y^2\)

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

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

\(\Leftrightarrow\left(2x-2y+1\right)\left(2x+2y+1\right)=1\)

Vậy \(\left(x;y\right)=\left(0;0\right);\left(-1;0\right)\)

Thay x=3 vào pt,ta được:

3^2+(m^2-2m)*3-9+12m=0

=>3m^2-6m+12m=0

=>3m^2+6m=0

=>m=0 hoặc m=-2

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

\(\dfrac{x+5}{x-5}=\dfrac{5}{x^2-5x}+\dfrac{1}{x}\)

\(\Leftrightarrow\dfrac{x+5}{x-5}=\dfrac{5}{x\left(x-5\right)}+\dfrac{1}{x}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x-5\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne5\end{matrix}\right.\)

Ta có : \(\dfrac{x+5}{x-5}=\dfrac{5}{x\left(x-5\right)}+\dfrac{1}{x}\)

\(\Leftrightarrow\dfrac{x\left(x+5\right)}{x\left(x-5\right)}=\dfrac{5}{x\left(x-5\right)}+\dfrac{x-5}{x\left(x-5\right)}\)

`=> x (x+5) = 5 +x-5`

`<=> x^2 +5x - 5-x+5=0`

`<=> x^2 +4x =0`

`<=> x(x+4)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-4\end{matrix}\right.\)

Vậy phương trình có nghiệm `x=-4`