phân tích các đa thức sau thành nhân tử:
a) \(\left(ab-1\right)^2+\left(a+b\right)^2\)
Ta thấy \(\left\{{}\begin{matrix}\left(ab-1\right)^2\ge0\\\left(a+b\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(ab-1\right)^2+\left(a+b\right)^2>0\) nên k phân tích thành nhân tử đc.
b) \(x^3+2x^2+2x+1\)
= \(x^3+x^2+x^2+x+x+1\)
= \(x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2+x+1\right)\)
c) \(x^3-4x^2+12x-27\)
= \(x^3-3x^2-x^2+3x+9x-27\)
= \(x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
= \(\left(x-3\right)\left(x^2-x+9\right)\)
d) \(x^4+2x^3+2x^2+2x+1\)
= \(x^4+x^3+x^3+x^2+x^2+x+x+1\)
= \(x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\)
= \(\left(x+1\right)\left(x^3+x^2+x+1\right)\)
= \(\left(x+1\right)\left[x^2\left(x+1\right)+\left(x+1\right)\right]\)
= \(\left(x+1\right).\left(x+1\right)\left(x^2+1\right)\)
= \(\left(x+1\right)^2\left(x^2+1\right)\)
a, \(\left(ab-1\right)^2+\left(a+b\right)^2\)
\(=a^2b^2-2ab+1+a^2+2ab+b^2\)
\(=a^2b^2+a^2+b^2+1=a^2.\left(b^2+1\right)+\left(b^2+1\right)\)
\(=\left(b^2+1\right).\left(a^2+1\right)\)
b, \(x^3+2x^2+2x+1\)
\(=x^3+x^2+x^2+x+x+1\)
\(=x^2.\left(x+1\right)+x.\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right).\left(x^2+x+1\right)\)
c, \(x^3-4x^2+12x-27\)
\(=x^3-3x^2-x^2+3x+9x-27\)
\(=x^2.\left(x-3\right)-x.\left(x-3\right)+9.\left(x-3\right)\)
\(=\left(x-3\right).\left(x^2-x+9\right)\)
d, \(x^4-2x^3+2x-1=x^4-x^3-x^3+x^2-x^2+x+x-1\)
\(=x^3.\left(x-1\right)-x^2.\left(x-1\right)-x.\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right).\left(x^3-x^2-x+1\right)\)
\(=\left(x-1\right).\left[x^2.\left(x-1\right)-\left(x-1\right)\right]\)
\(=\left(x-1\right).\left(x-1\right).\left(x^2-1\right)=\left(x-1\right)^2\left(x^2-1\right)\)
e, \(x^4+2x^3+2x^2+2x+1\)
\(=x^4+x^3+x^3+x^2+x^2+x+x+1\)
\(=x^3.\left(x+1\right)+x^2.\left(x+1\right)+x.\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right).\left(x^3+x^2+x+1\right)\)
\(=\left(x+1\right).\left[x^2.\left(x+1\right)+\left(x+1\right)\right]\)
\(=\left(x+1\right).\left(x+1\right).\left(x^2+1\right)=\left(x+1\right)^2\left(x^2+1\right)\)
Chúc bạn học tốt!!!
