\(p=\left(x+1\right)\left(x^2-x+1\right)+x-\left(x-1\right)\left(x^2+x+1\right)+2010\)\(=\left(x^3+1\right)+x-\left(x^3-1\right)+2010=x^3+1+x-x^3+1+2010=x+2012\)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 \(x=\dfrac{1}{5}\Rightarrow q=-80.\dfrac{1}{5}-1=-17\)

a) x(x-5)+2(x-5) = (x-5)(x+2)
b) (x-7)(x-2)

c) (x+2)(x^2+2x+4)+5y(x+2) = (x+2)(x^2+2x+4+5y) 

d) (x^2+8)^2 -16x^2 = (x^2+8-4x)(x^2+8+4x)

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

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

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

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

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

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

Ta có: \(16x+40=10\cdot3^2+5\cdot\left(1+2+3\right)\)

\(\Leftrightarrow16x+40=90+5\cdot6=120\)

hay x=5

x⁴+5x³-x²-16x+10=0

<=>x⁴+3x³-5x²+2x³+6x²-10x-5x²-15x+25=0

<=>x²(x²+3x-5)+2x(x²+3x-5)-5(x²+3x-5)=0

<=>(x²+2x-2)(x²+3x-5)=0

<=>x²+2x-2=0 hoặc x²+3x-5=0

\(\Leftrightarrow x=\pm\sqrt{3}-1\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{\sqrt{29}+3}{2}\\x=\frac{\sqrt{29}-3}{2}\end{cases}}\)