\(5,\dfrac{4}{x-2}+\dfrac{x}{x+1}-\dfrac{x^2-2}{\left(x-2\right)\left(x+1\right)}=0\left(dkxd:x\ne2;-1\right)\)

\(\Rightarrow4\left(x+1\right)+x\left(x-2\right)-x^2-2=0\)

\(\Rightarrow4x+4+x^2-2x-x^2-2=0\)

\(\Rightarrow2x+2=0\)

\(\Rightarrow x=-1\left(loai\right)\)

Vậy \(S=\varnothing\)

em c.ơn nhiều ạ 

\(3x-4x^2+6-8x>x^2+4x+4\)

\(\Leftrightarrow5x^2+9x-2>0\Leftrightarrow\left(5x-1\right)\left(x+2\right)>0\)

TH1 : \(\left\{{}\begin{matrix}5x-1>0\\x+2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{1}{5}\\x>-2\end{matrix}\right.\Leftrightarrow x>\dfrac{1}{5}\)

TH2 : \(\left\{{}\begin{matrix}5x-1< 0\\x+2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{1}{5}\\x< -2\end{matrix}\right.\Leftrightarrow x< -2\)

a, 5x(x-2) + (2-x)=0
⇔5x(x-2) - (x-2) =0
⇔(x-2)(5x-1)=0
\(\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy....
c, (x³ - x²) - 4x² + 8x -4 =0
⇔x³ - x² -4x² + 8x - 4=0
⇔x²(x-1) - 4x(x-1) +4(x-1) =0
⇔(x-1) (x-2)²=0
⇔\(\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy...
Phần b cậu có chép sai đề không?

.chỗ đó là giải phương trình hay PTĐTTNT vậy?

\(x(x-1)(x+1)-(x-3)(x^2+2x+9)\)

\(=x\left(x^2-1\right)-\left[x\left(x^2+2x+9\right)-3\left(x^2+2x+9\right)\right]\)

\(=x^3-x-\left(x^3+2x^2+9x-3x^2-6x-27\right)\)

\(=x^3-x-\left(x^3-x^2+3x-27\right)\)

\(=x^3-x-x^3+x^2-3x+27\)

\(=x^2-4x+27\)

#\(Toru\)

Đề đâu ah.