B1:
a) \(x^3-2x^2+x-2\)
= \(x^2\left(x-2\right)+\left(x-2\right)=\left(x-2\right)\left(x^2+1\right)\)
b) \(2x^3+3x^2-3x-2\)
= \(2x^3-2x^2+5x^2-5x+2x-2\)
= \(2x^2\left(x-1\right)+5x\left(x-1\right)+2\left(x-1\right)\)
= \(\left(x-1\right)\left(2x^2+5x+2\right)\)
= \(\left(x-1\right)\left(2x^2+4x+x+2\right)\)
= \(\left(x-1\right)\left[2x\left(x+2\right)+\left(x+2\right)\right]\)
= \(\left(x-1\right)\left(x+2\right)\left(2x+1\right)\)
c) \(5x^2+5y^2-x^2z+2xyz-y^2z-10xy\)
= \(5\left(x^2+2xy+y^2\right)+z\left(x^2+2xy+y^2\right)\)
= \(5\left(x+y\right)^2+z\left(x+y\right)^2\)
= \(\left(x+y\right)^2\left(5+z\right)\)
d) \(x^3-3x^2y+3xy^2-x+y-y^3\)
= \(\left(x-y\right)^3-\left(x-y\right)\)
= \(\left(x-y\right)\left[\left(x-y\right)^2-1\right]\)
= \(\left(x-y\right)\left(x-y-1\right)\left(x-y+1\right)\)
B2:
a) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
\(\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)
\(\left(2x-5\right)\left(2x+5-2x-7\right)=0\)
\(\left(2x-5\right).\left(-2\right)=0\)
\(\Rightarrow2x-5=0\Rightarrow x=\dfrac{5}{2}\)
b) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\left(x+3\right)\left(x^2-2x\right)=0\)
\(\left(x+3\right).x.\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=0\\x=2\end{matrix}\right.\)
c) \(2x^3+3x^2+2x+3=0\)
\(x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\left(2x+3\right)\left(x^2+1\right)=0\)
Ta thấy \(x^2+1>0\) với mọi x
\(\Rightarrow2x+3=0\Rightarrow x=\dfrac{-3}{2}\)
