Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}a.{(3x + 1)^2} - 4{(x - 2)^2}\\ = 9{x^2} + 6x + 1 - 4\left( {{x^2} - 4x + 4} \right)\\ = 9{x^2} + 6x + 1 - 4{x^2} + 16x - 16\\ = 5{x^2} + 22x - 15\\b.9{(2x + 3)^2} - 4{(x + 1)^2}\\ = 9.\left( {4{x^2} + 12x + 9} \right) - 4.\left( {{x^2} + 2x + 1} \right)\\ = 36{x^2} + 108x + 81 - 4{x^2} - 8x + 4\\ = 32{x^2} + 100x + 85\\c.4.{b^2}.{c^2} - {({b^2} + {c^2} - {a^2})^2}\\ = 4{b^2}{c^2} - \left( {{b^2} + {c^2} - {a^2}} \right)\left( {{b^2} + {c^2} - {a^2}} \right)\\ = 4{b^2}{c^2} - \left( {{b^4} + {b^2}{c^2} - {a^2}{b^2} + {b^2}{c^2} + {c^4} - {a^2}{c^2} - {a^2}{b^2} - {a^2}{c^2} + {a^4}} \right)\\ = 4{b^2}{c^2} - \left( {{a^4} + {b^4} + {c^4} + 2{b^2}{c^2} - 2{a^2}{b^2} - 2{a^2}{c^2}} \right)\\ = 4{b^2}{c^2} - {a^4} - {b^4} - {c^4} - 2{b^2}{c^2} + 2{a^2}{b^2} + 2{a^2}{c^2}\\ = 2\left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \right) - \left( {{a^4} + {b^4} + {c^4}} \right)\end{array}\)
