ĐKXĐ : x ≠ ±1/3

Ta có : \(\frac{3x-1}{6x+2}-\frac{3x+1}{2-6x}-\frac{6x}{9x^2-1}\)

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

\(=\frac{\left(3x-1\right)\left(3x-1\right)}{2\left(3x-1\right)\left(3x+1\right)}+\frac{\left(3x+1\right)\left(3x+1\right)}{2\left(3x-1\right)\left(3x+1\right)}-\frac{12x}{2\left(3x-1\right)\left(3x+1\right)}\)

\(=\frac{9x^2-6x+1}{2\left(3x-1\right)\left(3x+1\right)}+\frac{9x^2+6x+1}{2\left(3x-1\right)\left(3x+1\right)}-\frac{12x}{2\left(3x-1\right)\left(3x+1\right)}\)

\(=\frac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}\)

\(=\frac{18x^2-12x+2}{2\left(3x-1\right)\left(3x+1\right)}\)

\(=\frac{2\left(9x^2-6x+1\right)}{2\left(3x-1\right)\left(3x+1\right)}\)

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

\(\frac{x}{x-y}-\frac{1}{x-y}-\frac{1-y}{y-x}=\frac{x}{x-y}-\frac{1}{x-y}+\frac{y-1}{x-y}\)

\(=\frac{x-1+y-1}{x-y}=\frac{x+y-2}{x-y}\)

\(\frac{x}{x-y}-\frac{1}{x-y}-\frac{1-y}{y-x}\)

ĐKXĐ : \(\hept{\begin{cases}x,y\ne0\\x\ne y\end{cases}}\)

\(=\frac{x}{x-y}-\frac{1}{x-y}-\frac{y-1}{x-y}\)

\(=\frac{x-1-y+1}{x-y}\)

\(=\frac{x-y}{x-y}=1\)

\(\frac{x}{x-y}-\frac{1}{x-y}-\frac{1-y}{y-x}\)

ĐKXĐ : \(\hept{\begin{cases}x,y\ne0\\x\ne y\end{cases}}\)

\(=\frac{x}{x-y}-\frac{1}{x-y}-\frac{y-1}{x-y}\)

\(=\frac{x-1-y+1}{x-y}\)

\(=\frac{x-y}{x-y}=1\)

\(\frac{x}{x-y}-\frac{1}{x-y}-\frac{1-y}{y-x}=\frac{x-1}{x-y}+\frac{1-y}{x-y}=\frac{x-1+1-y}{x-y}=\frac{x-y}{x-y}=1\)

\(ĐKXĐ:\hept{\begin{cases}x\ne-1\\x\ne1\end{cases}}\)

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

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

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

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

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

\(ĐKXĐ:\hept{\begin{cases}x\ne0\\y\ne0\\x\ne y\end{cases}}\)

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

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

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

\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)          (điều kiện: \(x;y\ne0\); \(x\ne\pm2y\))

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

\(=\frac{2}{x+2y}+\frac{1}{x-2y}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)

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

\(=\frac{3x-2y+4}{\left(x-2y\right)\left(x+2y\right)}\)

3x^2-4x^2+2

=-x^2+2

= - ( x^2 -2 )

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

3x³ - 4x² + x

= x( 3x² - 4x + 1 )

= x( 3x² - 3x - x + 1 )

= x[ ( 3x² - 3x ) - ( x - 1 ) ]

= x[ 3x( x - 1 ) - ( x - 1 ) ]

= x( x - 1 )( 3x - 1 )

3x ³ - 4x ² + x

= x( 3x ² - 4x + 1 )

= x( 3x ² - 3x - x + 1 )

= x[ ( 3x ² - 3x ) - ( x - 1 ) ]

= x[ 3x( x - 1 ) - ( x - 1 ) ]

= x( x - 1 )( 3x - 1 )

Ta có : \(\left(x^4-2x^3+2x+a\right)\div\left(x-1\right)^2\)

\(\Rightarrow\left(x^4-2x^3+2x+a\right)\div\left(x^2-2x+1\right)\)

Ta đặt phép chia :

x^4 - 2x^3 + 2x + a x^2 - 2x + 1 x^2 - 1 x^4 - 2x^3 + x^2 -x^2 + 2x + a -x^2 + 2x - 1 a + 1

Để đa thức trên chia hết thì \(a+1=0\)

                                          \(\Rightarrow a=-1\)

Vậy \(a=-1\) thì \(\left(x^4-2x^3+2x+a\right)⋮\left(x-1\right)^2\)