Bài 1.
a) \({x^4} + 2{x^2}y + {y^2} = \left( {{x^2} + {y^2}} \right).\)
b) \({\left( {2a + b} \right)^2} - {\left( {2b + a} \right)^2} \)
\(= \left[ {\left( {2a + b} \right) + \left( {2b + a} \right)} \right]\left[ {\left( {2a + b} \right) - \left( {2b + a} \right)} \right]\)
\( = \left( {3a + 3b} \right)\left( {a - b} \right) = 3\left( {a + b} \right)\left( {a - b} \right).\)
c) \(\left( {8{a^3} - 27{b^3}} \right) - 2a\left( {4{a^2} - 9{b^2}} \right)\)
\( = \left( {2a - 3b} \right)\left( {4{a^2} + 6ab + 9{b^2}} \right) - 2a\left( {2a - 3b} \right)\left( {2a + 3b} \right)\)
\( = \left( {2a - 3b} \right)\left( {4{a^2} + 6ab + 9{b^2} - 4{a^2} - 6ab} \right)\)
\(= 9{b^2}\left( {2a - 3b} \right).\)
Bài 2.
\({x^2} - 36 = 0\)
\(\Rightarrow \left( {x + 6} \right)\left( {x - 6} \right) = 0\)
\( \Rightarrow x + 6 = 0\) hoặc \(x - 6 = 0 \)
\(\Rightarrow x = - 6\) hoặc \(x = 6.\)
Bài 3. Ta có:
\({\left( {5n - 2} \right)^2} - {\left( {2n - 5} \right)^2} \)
\(= \left( {5n - 2 + 2n - 5} \right)\left( {5n - 2 - 2n + 5} \right)\)
\( = \left( {7n - 7} \right)\left( {3n + 3} \right) \)
\(= 21\left( {n - 1} \right)\left( {n + 1} \right)\; \vdots\; 21\) , với mọi n thuộc \(\mathbb Z\)
