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

\(=x\left(x+3\right)\left[x\left(x+2\right)+\left(x+2\right)\right]=x\left(x+3\right)\left(x+1\right)\left(x+2\right)\)

2) \(x^4+2012x^2+2011x+2012\)

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

\(=x\left(x^3-1\right)+2012\left(x^2+x+1\right)\)

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

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

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

= x³ + y³ + z³ + 3x²yz + 3xy²z + 3xyz² - x³ -y³ - z³

=3x²yz + 3xy²z + 3xyz²

= 3xyz( x + y + z)

b.

x^4+2012x^2+2012x-x+2012=

(x^4-x)+2012(x^2+x+1)=

x(x-1)(x^2+x+1)+2012(x^2+x+1)=

(x+2012)(x^2+x+1)

làm sao ra vậy

a) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)

\(=\left[\left(x+y\right)^3+z^3+3.\left(x+y\right).z.\left(x+y+z\right)\right]-x^3-y^3-z^3\)

\(=\left[x^3+y^3+3xy.\left(x+y\right)+z^3+3\left(x+y\right).z.\left(x+y+z\right)\right]-x^3-y^3-z^3\)

\(=3xy\left(x+y\right)+3\left(x+y\right)z.\left(x+y+z\right)\)

\(=3.\left(x+y\right)\left(xy+zx+zy+z^2\right)\)

\(=3.\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

b) \(x^4+2012x^2+2011x+2012\)

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

\(=x.\left(x^3-1\right)+2012.\left(x^2+x+1\right)\)

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

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

x⁴+2012x²+2012x+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)

\(a\text{)}\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=\left(x+y+z-x\right)\left[\left(x+y+z\right)^2+x\left(x+y+z\right)+x^2\right]-\left(y^3+z^3\right)\)

\(=\left(y+z\right)\left(3x^2+y^2+z^2+3xy+3xz+2yz\right)-\left(y+z\right)\left(y^2-yz+z^2\right)\)

\(=\left(y+z\right)\left(3x^2+y^2+z^2+3xy+3xz+2yz-y^2+yz-z^2\right)\)

\(=\left(y+z\right)\left(3x^2+3xy+3yz+3xz\right)\)

\(=3\left(y+z\right)\left(x^2+xy+yz+xz\right)\)

\(=3\left(y+z\right)\left(x+y\right)\left(x+z\right)\)

\(b\text{)}x^4+2012x^2+2011x+2012\)

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

\(=x\left(x^3-1\right)+2012\left(x^2+x+1\right)\)

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

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

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

=(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+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)

     \(2012x^2-x-2013=0\)

\(\Rightarrow2012x^2+2012x-2013x-2013=0\)

\(\Rightarrow2012x\left(x+1\right)-2013\left(x+1\right)=0\)

\(\Rightarrow\left(2012x-2013\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2012x-2013=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{2013}{2012}\\x=-1\end{cases}}}\)

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