a. (3x+1)^2 – 4(x-2)^2 b. 9(2x+3)^2 – 4(x+1)^2 c. 4*b^2*c^2 – (b^2+c^2-a^2)^2

a. (3x+1)^2 – 4(x-2)^2
b. 9(2x+3)^2 – 4(x+1)^2
c. 4*b^2*c^2 – (b^2+c^2-a^2)^2

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  1. Đáp án:

    Giải thích các bước giải:

    \(\begin{array}{l}a.{(3x + 1)^2} – 4{(x – 2)^2}\\ = 9{x^2} + 6x + 1 – 4\left( {{x^2} – 4x + 4} \right)\\ = 9{x^2} + 6x + 1 – 4{x^2} + 16x – 16\\ = 5{x^2} + 22x – 15\\b.9{(2x + 3)^2} – 4{(x + 1)^2}\\ = 9.\left( {4{x^2} + 12x + 9} \right) – 4.\left( {{x^2} + 2x + 1} \right)\\ = 36{x^2} + 108x + 81 – 4{x^2} – 8x + 4\\ = 32{x^2} + 100x + 85\\c.4.{b^2}.{c^2} – {({b^2} + {c^2} – {a^2})^2}\\ = 4{b^2}{c^2} – \left( {{b^2} + {c^2} – {a^2}} \right)\left( {{b^2} + {c^2} – {a^2}} \right)\\ = 4{b^2}{c^2} – \left( {{b^4} + {b^2}{c^2} – {a^2}{b^2} + {b^2}{c^2} + {c^4} – {a^2}{c^2} – {a^2}{b^2} – {a^2}{c^2} + {a^4}} \right)\\ = 4{b^2}{c^2} – \left( {{a^4} + {b^4} + {c^4} + 2{b^2}{c^2} – 2{a^2}{b^2} – 2{a^2}{c^2}} \right)\\ = 4{b^2}{c^2} – {a^4} – {b^4} – {c^4} – 2{b^2}{c^2} + 2{a^2}{b^2} + 2{a^2}{c^2}\\ = 2\left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2}} \right) – \left( {{a^4} + {b^4} + {c^4}} \right)\end{array}\)

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