1. Tính:
a)(x+4) (x^2 - 4x+16)=(x+4)(x2-4x+42)=x3+64
b)(x-3y) (x^2 + 3xy+3y^2)=x3-(3y)3
c)(x-3)^2 -(x+3)^3=(x2-6x+9)-(x3+9x2+27x+27)=-8x2-33x-18-x3
d)(x^2-1)^3 - (x^2 +1)^3=(x2-1-x2-1)[(x2-1)2+(x2-1)(x2+1)+(x2+1)2]
=-2(x4-2x2+1+x4-1+x4+2x2+1)
=-2(3x4+2)=-6x4-4
a/ \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=0\)
\(\Leftrightarrow x^3-2x^2+4x+2x^2-4x+8-x^3-2x=0\)
\(\Leftrightarrow-2x=-8\)
\(\Leftrightarrow x=4\)
Vậy .....................
b/ \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow x^6-3x^4+3x^2-1-\left(x^6-x^4+x^4-x^2+x^2-1\right)=0\)
\(\Leftrightarrow x^6-3x^4+3x^2-1-x^6+1=0\)
\(\Leftrightarrow-3x^4+3x^2=0\)
\(\Leftrightarrow3x^2\left(-x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2=0\\-x^2+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x^2=-1\Rightarrow x^2=1\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\end{matrix}\right.\)
Vậy pt có 3 nghiệm là \(\left\{{}\begin{matrix}x=-1\\x=0\\x=1\end{matrix}\right.\)
