\(\left(x^3-2x^2+x+4\right):\left(x+1\right)\)

\(=\left(x^3-3x^2+4x+x^2-3x+4\right):\left(x+1\right)\)

\(=\left[x\left(x^2-3x+4\right)+\left(x^2-3x+4\right)\right]:\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-3x+4\right):\left(x+1\right)\)

\(=x^2-3x+4\)

\(\left[\left(3-x\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2\right]:\left(x^2-6x+9\right)=\left[\left(3-x\right)^5-7\left(3-x\right)^4-4\left(3-x\right)^2\right]:\left(3-x\right)^2=\left(3-x\right)^2\left[\left(3-x\right)^3-7\left(3-x\right)^2-4\right]:\left(3-x\right)^2=\left(3-x\right)^3-7\left(3-x\right)^2-4=27-27x+9x^2-x^3-63+42x-7x^2-4=-x^3+2x^2+15x-40\)

\(\dfrac{\left(3-x\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2}{x^2-6x+9}\)

\(=\dfrac{-\left(x-3\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2}{\left(x-3\right)^2}\)

\(=-\left(x-3\right)^3-7\left(x-3\right)^2-4\)

Vậy \(\left(2x^4+2x^3+3x^2-5x-20\right):\left(x^2+x+4\right)=2x^2-5\)

`8x^4y^5z^3 : 2x^3y^4z`

`= 4xyz^2`.

`(-x^4+2x-3x^2):(x-2)`

`=[-x(x^3+3x-2)]:(x-2)`

`=[-x(x^3-2x^2+2x^2-4x+7x-14+12)]:(x-2)`

`={-x[x^2(x-2)+2x(x-2)+7(x-2)]-12x+24-24}:(x-2)`

`=[-x(x-2)(x^2+2x+7)-12(x-2)-24]:(x-2)`

`=-x(x^2+2x+7)-12` và dư `-24`

`=-x^3-2x^2-7x-12` và dư `-24`

\(\dfrac{-x^4-3x^2+2x}{x-2}\)

\(=\dfrac{-x^4+2x^3-2x^3+4x^2-7x^2+14x-12x+24-24}{x-2}\)

\(=-x^3-2x^2-7x-12-\dfrac{24}{x-2}\)

\(=\left(3x^4-3x^3+x^3-x^2+8x^2-8x+9x-9\right):\left(x-1\right)\\ =\left(x-1\right)\left(3x^3+x^2+8x+9\right):\left(x-1\right)\\ =3x^3+x^2+8x+9\)

a: \(=\dfrac{5\left(x+2\right)}{10xy^2}\cdot\dfrac{12x}{x+2}=\dfrac{60x}{10xy^2}=\dfrac{6}{y^2}\)

b: \(=\dfrac{x-4}{3x-1}\cdot\dfrac{3\left(3x-1\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{3}{x+4}\)

c: \(=\dfrac{2\left(2x+1\right)}{\left(x+4\right)^2}\cdot\dfrac{\left(x+4\right)}{3\left(x+3\right)}=\dfrac{2\left(2x+1\right)}{3\left(x+3\right)\left(x+4\right)}\)

d: \(=\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\cdot\dfrac{x+1}{x-1}=\dfrac{5}{3}\)

help me