Bài 5. Hãy mở các dấu ngoặc sau:
a) (4n2 – 6mn + 9m2)(2n + 3m)
b) (7 + 2b)(4b2 – 4b + 49);
c) (25a2 + 10ab + 4b2)(5a – 2b);
d)(x2 + x + 2)(x2 – x – 2).
Bài 5. Hãy mở các dấu ngoặc sau:
a) (4n2 – 6mn + 9m2)(2n + 3m)
b) (7 + 2b)(4b2 – 4b + 49);
c) (25a2 + 10ab + 4b2)(5a – 2b);
d)(x2 + x + 2)(x2 – x – 2).
\[\begin{array}{l}
a)\,\,\,\left( {4{n^2} – 6mn + 9{m^2}} \right)\left( {2n + 3m} \right) = {\left( {2n} \right)^3} + {\left( {3m} \right)^3} = 8{n^3} + 27{m^3}.\\
b)\,\,\left( {7 + 2b} \right)\left( {4{b^2} – 4b + 49} \right) = {7^3} + {\left( {2b} \right)^3} = 343 + 8{b^3}.\\
c)\,\,\left( {25{a^2} + 10ab + 4{b^2}} \right)\left( {5a – 2b} \right) = {\left( {5a} \right)^3} – {\left( {2b} \right)^3} = 125{a^3} – 8{b^3}.\\
d)\,\,\,\left( {{x^2} + x + 2} \right)\left( {{x^2} – x – 2} \right) = \left[ {{x^2} + \left( {x + 2} \right)} \right]\left[ {{x^2} – \left( {x + 2} \right)} \right]\\
= {\left( {{x^2}} \right)^2} – {\left( {x + 2} \right)^2} = {x^4} – \left( {{x^2} + 4x + 4} \right) = {x^4} – {x^2} – 4x – 4.
\end{array}\]