Đáp án:

$\begin{array}{l}
a){\left( {ab - 1} \right)^2} + {\left( {a + b} \right)^2}\\
 = {a^2}{b^2} - 2ab + 1 + {a^2} + 2ab + {b^2}\\
 = {a^2}{b^2} + 1 + {a^2} + {b^2}\\
 = {a^2}\left( {{b^2} + 1} \right) + \left( {{b^2} + 1} \right)\\
 = \left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\\
c){x^3} - 4{x^2} + 12x - 27\\
 = {x^3} - 27 + \left( { - 4{x^2} + 12x} \right)\\
 = \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) - 4x\left( {x - 3} \right)\\
 = \left( {x - 3} \right)\left( {{x^2} + 3x + 9 - 4x} \right)\\
 = \left( {x - 3} \right)\left( {{x^2} - x + 9} \right)\\
b){x^3} + 2{x^2} + 2x + 1\\
 = {x^3} + 2{x^2} + x + x + 1\\
 = x\left( {{x^2} + 2x + 1} \right) + \left( {x + 1} \right)\\
 = x{\left( {x + 1} \right)^2} + \left( {x + 1} \right)\\
 = \left( {x + 1} \right)\left( {x\left( {x + 1} \right) + 1} \right)\\
 = \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\
d){x^4} - 2{x^3} + 2x - 1\\
 = {x^4} - 2{x^3} + {x^2} - {x^2} + 2x - 1\\
 = {x^2}\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)\\
 = \left( {{x^2} - 2x + 1} \right)\left( {{x^2} - 1} \right)\\
 = {\left( {x - 1} \right)^2}\left( {x - 1} \right)\left( {x + 1} \right)\\
 = {\left( {x - 1} \right)^3}\left( {x + 1} \right)\\
e){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\
 = {x^4} + 2{x^3} + {x^2} + {x^2} + 2x + 1\\
 = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\
 = \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right)\\
 = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right)
\end{array}$

Đáp án:

a)(ab−1)2+(a+b)2

=a2b2−2ab+1+a2+2ab+b2

=a2b2+1+a2+b2=a2(b2+1)+(b2+1) = (a2+1)(b2+1)

c)x3−4x2+12x−27

=x3−27+(−4x2+12x)

=(x−3)(x2+3x+9)−4x(x−3)

=(x−3)(x2+3x+9−4x)

=(x−3)(x2−x+9)

b)x3+2x2+2x+1

=x3+2x2+x+x+1

=x(x2+2x+1)+(x+1)

=x(x+1)2+(x+1)

=(x+1)(x(x+1)+1)

=(x+1)(x2+x+1)

d)x4−2x3+2x−1

=x4−2x3+x2−x2+2x−1

=x2(x2−2x+1)−(x2−2x+1)

=(x2−2x+1)(x2−1)

=(x−1)2(x−1)(x+1)

=(x−1)3(x+1)

e)x4+2x3+2x2+2x+1

=x4+2x3+x2+x2+2x+1

=x2(x2+2x+1)+(x2+2x+1)

=(x2+2x+1)(x2+1)

=(x+1)2(x2+1)