\(p=\left(x+1\right)\left(x^2-x+1\right)+x-\left(x-1\right)\left(x^2+x+1\right)+2010\)\(=\left(x^3+1\right)+x-\left(x^3-1\right)+2010=x^3+1+x-x^3+1+2010=x+2012\)Với \(x=-2010\Rightarrow p=-2010+2012=2\)

\(q=16x\left(4x^2-5\right)-\left(4x+1\right)\left(16x^2-4x+1\right)=64x^3-80x-64x^3-1=-80x-1\)Với \(x=\dfrac{1}{5}\Rightarrow q=-80.\dfrac{1}{5}-1=-17\)

giải

5x-(4-2x+x^2)(x+2)+x(x-1)(x+1)=0

5x-(4x+8-2x^2-4x+x^3+2x^2)+x(x^2-1)=0

5x-4x-8+2x^2+4x-x^3-2x^2+x^3-1x=0

(5x-4x+4x-1x)+(-8)+(2x^2-2x^2)+(-x^3+x^3)=0

4x+(-8)=0

4x=0+8

4x=8

x=8:4

x=2

D)(4x+1)(16x^2-4x+1)-16x(4x^2-5)=17

64x^3-16x^2+4x+16x^2-4x+1-64x^3+80x=17

80x+1=17

80x=17-1

80x=16

x=1/5

\(5x-\left(4-2x+x^2\right)\left(x+2\right)+x\left(x-1\right)\left(x+1\right)=0\)

\(\Rightarrow5x-\left(4x-2x^2+x^3+8-4x+2x^2\right)+\left(x^2-x\right)\left(x+1\right)=0\)

\(\Rightarrow5x-\left(4x-2x^2+x^3+8-4x+2x^2\right)+\left(x^3+x-x^2-x\right)=0\)

\(\Rightarrow5x-4x+2x^2-x^3-8+4x-2x^2+x^3+x-x^2-x=0\)

\(\Rightarrow4x-8=0\Rightarrow4x=8\Rightarrow x=2\)

(x - 1)² - (x + 2)(x - 5) + (4x - 1)² = (-x - 1)(2 - 16x)

=> x² - 2x + 1 - x² + 5x - 2x + 10 + 16x² - 8x + 1 = -2x + 16x² -2 + 16x

=> 16x² - 7x + 12 = 14x - 2  + 16x²

=> -21x = -14

=> 21x = 14

=> x = 2/3

\(\left(x-1\right)^2-\left(x+2\right)\left(x-5\right)+\left(4x-1\right)^2=\left(-x-1\right)\left(2-16x\right)\)

\(\Rightarrow x^2-2x+1-x^2+3x+10+16x^2-8x+1-16x^2-14x+2=0\)

\(\Rightarrow\left(x^2-x^2+16x^2-16x^2\right)+\left(-2x+3x-8x-14x\right)+\left(1+10+1+2\right)=0\)

\(\Rightarrow-21x+14=0\)

\(\Rightarrow-21x=-14\)

\(\Rightarrow x=\frac{2}{3}\)

\(x^2-2x+1-\left(x^2-3x-10\right)+16x^2-8x+1=16x^2+14x-2\)    

\(x^2-2x+1-x^2+3x+10+16x^2-8x+1-16x^2-14x+2=0\)         

\(-21x-14=0\) 

\(-21x=14\)       

\(x=\frac{-2}{3}\)

64x^3 + 1 - 64x^3 + 80x =17

80x                              =16

   x                               =3/10

64x^3 + 1 -64x^3 + 80x = 17

80x                             = 16

   x                              = 3/10

64x^3 + 1 - 64x^3 + 80x = 17

80x                               = 16

   x                                = 3/10

b) \(\left(4x+1\right)\left(16x^2-4x+1\right)-16x\left(4x^2-5\right)=17\)

\(\Leftrightarrow64x^3+1-64x^3+80x=17\)

\(\Leftrightarrow80x=16\)

\(\Leftrightarrow x=\frac{1}{5}\)

Bài 1: 

c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)

Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)

\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)

Suy ra: \(-12x-3=8x-2-6x-8\)

\(\Leftrightarrow-12x-3-2x+10=0\)

\(\Leftrightarrow-14x+7=0\)

\(\Leftrightarrow-14x=-7\)

\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)