Ta có :

x = 2012

x - 1 = 2011

P(x) = x²⁰¹² - 2011x²⁰¹¹ - 2011x²⁰¹⁰ - .... - 2011x² - 2011x - 1

P(x) = x²⁰¹² - (x - 1)x²⁰¹¹ - (x - 1)x²⁰¹⁰ - ..... - (x - 1)x² - (x - 1)x - 1

P(x) = x²⁰¹² - x²⁰¹² + x²⁰¹¹ -  x²⁰¹¹ + x²⁰¹⁰ - ...... - x³ + x² - x² + x - 1

P(x) = x - 1

P(2012) = 2012 - 1 = 2011

Thay 2011 = x - 1 vào P(2012) rồi nhân vào nó sẽ tự triệt tiêu hết

Phải là 2011x+1chứ ko phải -1

Thay xyz = 2011 vào N được : 

\(N=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}=\frac{xy.xz}{xy\left(z+xz+1\right)}+\frac{y}{y\left(z+xz+1\right)}+\frac{z}{z+xz+1}\)

\(=\frac{xz}{z+xz+1}+\frac{1}{z+xz+1}+\frac{z}{z+xz+1}=\frac{z+xz+1}{z+xz+1}=1\)

b: 5x^2+5y^2+8xy-2x+2y+2=0

=>4x^2+8xy+4y^2+x^2-2x+1+y^2+2y+1=0

=>(x-1)^2+(y+1)^2+(2x+2y)^2=0

=>x=1 và y=-1

M=(1-1)^2015+(1-2)^2016+(-1+1)^2017=1

=(x4−x3+2011x2)+

(x3−x2+2011x)+(x2−x+2011)

=x2(x2−x+2011)+x(x2−x+2011)+(x2−x+2011)

=(x2+x+1)(x2−x+2011)

=(x4−x3+2011x2)+(x3−x2+2011x)+(x2−x+2011)

=x2(x2−x+2011)+x(x2−x+2011)+(x2−x+2011)

=(x2+x+1)(x2−x+2011)

x3−x2+2011x)+(x2−x+2011)

=x2(x2−x+2011)+x(x2−x+2011)+(x2−x+2011)=(x2+x+1)(x2−x+2011)

x⁴+2012x²+2011x+2012

=(x⁴-x)+(2012x²+2012x+2012)

=x(x³-1)+2012(x²+x+1)

=x(x-1) (x²+x+1) + 2012 (x²+x+1)

=(x²+x+1) [x(x-1)+2012]

=(x²+x+1) (x²-x+2012)

\(x^4+2012x^2+2011x+2012\)

\(=x^4-x+2012x^2+2012x+2012\)

\(=x.\left(x-1\right)\left(x^2+x+1\right)+2012.\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2-x+2012\right)\)

Chọn A

x⁴+2011x²+2010x+2011

=(x⁴+x³+x²)+(2011x²+2011x+2011)-(x³+x²+x)

=x²(x²+x+1)+2011(x²+x+1)-x(x²+x+1)

=(x²+x+1)(x²+2011-x)

x⁴+2011x²+2010x+2011=x⁴-x+2011x²+2011x+2011

                                    =x(x³-1)+2011(x²+x+1)

                                    =x(x- 1)(x²+x+1)+2011(x²+x+1)

                                   =(x²+x+1)[x(x-1)+2011]

                                    =(x²+x+1)(x²-x+2011)