\(\frac{x+2}{x}+\frac{2x-1}{2-x}-\frac{x-8}{x^2-2x}\)

\(=\frac{x+2}{x}-\frac{2x-1}{x-2}-\frac{x-8}{x\left(x-2\right)}\)

\(=\frac{\left(x-2\right)^2}{x\left(x-2\right)}-\frac{x\left(2x-1\right)}{x\left(x-2\right)}-\frac{x-8}{x\left(x-2\right)}\)

\(=\frac{x^2-4x+4-2x^2+x-x+8}{x\left(x-2\right)}=\frac{-x^2-4x+12}{x\left(x-2\right)}\)

\(=\frac{\left(x+6\right)\left(x-2\right)}{x\left(x-2\right)}=\frac{x+6}{x}\)

Bài làm 

\(\frac{xy}{x^2-y^2}+\frac{x^2+2x}{x^2-y^2}=\frac{xy+x^2+2x}{x^2-y^2}=\frac{x\left(x+2+y\right)}{\left(x-y\right)\left(x+y\right)}\)

Học tốt 

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

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

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

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=2;\pm1\)

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

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

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

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

\(\Leftrightarrow\hept{\begin{cases}x-2=0\\x-1=0\\x+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\x=1\\x=-1\end{cases}}}\)

Vậy \(x=2;x=1;x=-1\)

\(x^2+6x=0\)

\(x.\left(x+6\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x+6=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=-6\end{cases}}}\)

Vậy \(x=0;x=-6\)

\(\left(x-4\right)\left(x+4\right)-x\left(x-2\right)=0\)

\(x^2-16-x^2+2x=0\)

\(2x-16=0\)

\(2.\left(x-8\right)=0\)

\(x-8=0\)

\(x=8\)

Vậy \(x=8\)

\(x^2+6x=0\Leftrightarrow x\left(x+6\right)=0\Leftrightarrow x=0;-6\)

\(\left(x-4\right)\left(x+4\right)-x\left(x-2\right)=0\)

\(\Leftrightarrow x^2-16-x^2+2x=0\Leftrightarrow-16+2x=0\Leftrightarrow x=8\)

a,\(x^2y-4y=y\left(x^2-4\right)=y\left(x-2\right)\left(x+2\right)\)

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

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

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

c,\(5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)\)

                                            \(=\left(5x-1\right)\left(x+y\right)\)

x²y - 4y = y( x² - 4 ) = y( x - 2 )( x + 2 )

x² - y² - 2x + 1 = ( x² - 2x + 1 ) - y² = ( x - 1 )² - y² = ( x - 1 - y )( x - 1 + y )

5x² + 5xy - x - y = ( 5x² + 5xy ) - ( x + y ) = 5x( x + y ) - ( x + y ) = ( x + y )( 5x - 1 )

+)\(\frac{x+2}{2x-4}-\frac{4x}{x^2-4}=\frac{x+2}{2\left(x-2\right)}-\frac{4x}{\left(x-2\right)\left(x+2\right)}\)

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

\(=\frac{x^2+4x+4-8x}{2\left(x-2\right)\left(x+2\right)}==\frac{x^2-4x+4}{2\left(x-2\right)\left(x+2\right)}\)

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

+) \(\frac{2x+6}{3x^2-x}-\frac{x+3}{1-3x}=\frac{2x+6}{x\left(3x-1\right)}-\frac{-x-3}{3x-1}\)

\(=\frac{2x+6}{x\left(3x-1\right)}-\frac{x\left(-x-3\right)}{x\left(3x-1\right)}\)

\(=\frac{2x+6+x^2+3x}{x\left(3x-1\right)}=\frac{x^2+5x+6}{x\left(3x-1\right)}\)

a)\(\frac{x+2}{2x-4}-\frac{4x}{x^2-4}\)

\(=\frac{x+2}{2\left(x-2\right)}-\frac{4x}{\left(x-2\right)\left(x+2\right)}\)

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

\(=\frac{x^2+4x+4-8x}{2\left(x+2\right)\left(x-2\right)}\)

\(=\frac{x^2-4x+4}{2\left(x+2\right)\left(x-2\right)}\)

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

\(=\frac{x-2}{2\left(x+2\right)}\)

b)\(\frac{2x+6}{3x^2-x}-\frac{x+3}{1-3x}\)

\(=\frac{2x+6}{x\left(3x-1\right)}-\frac{-x-3}{3x-1}\)

\(=\frac{2x+6+x\left(x+3\right)}{x\left(3x-1\right)}\)

\(=\frac{2x+6+x^2+3x}{x\left(3x-1\right)}\)

\(=\frac{x^2+5x+6}{x\left(3x-1\right)}\)

#Hoctot

\(\frac{x^3+x^2+x+1}{3x^2+6x+3}=\frac{x^2\left(x+1\right)+\left(x+1\right)}{3x^2+3x+3x+3}\)

\(=\frac{\left(x^2+1\right)\left(x+1\right)}{3x\left(x+1\right)+3\left(x+1\right)}=\frac{\left(x^2+1\right)\left(x+1\right)}{\left(3x+3\right)\left(x+1\right)}\)

\(=\frac{\left(x^2+1\right)\left(x+1\right)}{3\left(x+1\right)^2}=\frac{x^2+1}{3\left(x+1\right)}\)

\(\frac{x^3+x^2+x+1}{3x^2+6x+3}=\frac{x^2.\left(x+1\right)+\left(x+1\right)}{3.\left(x^2+2x+1\right)}=\frac{\left(x+1\right)\left(x^2+1\right)}{3.\left(x+1\right)^2}=\frac{x^2+1}{3.\left(x+1\right)}\)