Bài 1.
a) \(\left( {{a^3} - {b^3}} \right) + {\left( {a - b} \right)^2} \)
\(= \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) + \left( {a - b} \right)\left( {a - b} \right)\)
\(= \left( {a - b} \right)\left( {{a^2} + ab + {b^2} + a - b} \right).\)
b) \({\left( {{x^2} + 1} \right)^2} - 4{x^2}\)
\(= {\left( {{x^2} + 1} \right)^2} - {\left( {2x} \right)^2} \)
\(= \left( {{x^2} + 1 - 2x} \right)\left( {{x^2} + 1 + 2x} \right)\)
\( = {\left( {x - 1} \right)^2}{\left( {x + 1} \right)^2}.\)
c) \(\left( {{y^3} + 8} \right) + \left( {{y^2} - 4} \right) \)
\(= \left( {{y^3} + {2^3}} \right) + \left( {{y^2} - {2^2}} \right)\)
\( = \left( {y + 2} \right)\left( {{y^2} - 2y + 4} \right) + \left( {y + 2} \right)\left( {y - 2} \right)\)
\( = \left( {y + 2} \right)\left( {{y^2} - 2y + 4 + y - 2} \right) \)
\(= \left( {y + 2} \right)\left( {{y^2} - y + 2} \right).\)
Bài 2.
a) \({\left( {3x - 5} \right)^2} - {\left( {x + 1} \right)^2} \)
\(= \left( {3x - 5 + x + 1} \right)\left( {3x - 5 - x - 1} \right)\)
\( = \left( {4x - 4} \right)\left( { - 2x - 4} \right) \)
\(= 8\left( {x - 1} \right)\left( {x - 3} \right)\)
Vậy \(\left( {x - 1} \right)\left( {x - 3} \right) = 0\)
\(\Rightarrow x - 1 = 0\) hoặc \(x - 3 = 0\)
\( \Rightarrow x = 1\) hoặc \(x = 3.\)
b) \({\left( {5x - 4} \right)^2} - 49{x^2} \)
\(= {\left( {5x - 4} \right)^2} - {\left( {7x} \right)^2} \)
\(= \left( {5x - 4 + 7x} \right)\left( {5x - 4 - 7x} \right)\)
\( = \left( {12x - 4} \right)\left( { - 2x - 4} \right) \)
\(= - 8\left( {3x - 1} \right)\left( {x + 2} \right)\)
Vậy \(\left( {3x - 1} \right)\left( {x + 2} \right) = 0 \)
\(\Rightarrow 3x - 1 = 0\) hoặc \(x + 2 = 0\)
\(\Rightarrow x ={1\over 3}\) hoặc \(x = -2 \)
