b1:phân tích đa thức thành nhân tử a)4x^2-12x+9 b)4x^2+4x+1 c)1+12x+36x^2 d)(3x+1)^2-4(x-2)^2 e)9(2x+3)^2-4(x+1)^2 f)4b^2c-(b^2+c^2-a^2)^2 GIÚP VỚI Ạ

Đáp án+Giải thích các bước giải:

a)4𝑥²−12𝑥+9

=4𝑥²−6𝑥−6𝑥+9

=2𝑥(2𝑥−3)−3(2𝑥−3)

=(2𝑥−3)(2𝑥−3)

=(2𝑥−3)²

b)4x²+4x+1

=4𝑥²+2𝑥+2𝑥+1

 =2𝑥(2𝑥+1)+2𝑥+1

=(2𝑥+1)(2𝑥+1)

=(2𝑥+1)²

c)1+12x+36x²

= 36x² + 12x + 1

=36𝑥²+6𝑥+6𝑥+1

=6𝑥(6𝑥+1)+6𝑥+1

=(6𝑥+1)(6𝑥+1)

=(6𝑥+1)²

d)(3x+1)²-4(x-2)²

=(3𝑥+1)(3𝑥+1)−4(𝑥−2)²

=3𝑥(3𝑥+1)+1(3𝑥+1)−4(𝑥−2)²

=9𝑥²+3𝑥+1(3𝑥+1)−4(𝑥−2)²

=9𝑥²+3𝑥+3𝑥+1−4(𝑥−2)²

=9𝑥²+6𝑥+1−4(𝑥−2)²

=9𝑥²+6𝑥+1−4(𝑥−2)(𝑥−2)

=9𝑥²+6𝑥+1−4(𝑥(𝑥−2)−2(𝑥−2)

=9𝑥²+6𝑥+1−4(𝑥²−2𝑥−2(𝑥−2)

=9𝑥²+6𝑥+1−4(𝑥²−2𝑥−2𝑥+4)

=9𝑥²+6𝑥+1−4(𝑥²−4𝑥+4)

=9𝑥²+6𝑥+1−4𝑥²+16𝑥−16

=9𝑥²+6𝑥−15−4𝑥²+16𝑥

=5𝑥²+6𝑥−15+16𝑥

=5𝑥²+22𝑥−15

e)9(2x+3)²-4(x+1)²

=9(2𝑥+3)(2𝑥+3)−4(𝑥+1)²

=9(4𝑥²+12𝑥+9)−4(𝑥+1)²

=36𝑥²+108𝑥+81−4(𝑥+1)²

=36𝑥²+108𝑥+81−4(𝑥²+𝑥+𝑥+1)

=32𝑥²+108𝑥+77−8𝑥

=32𝑥²+100𝑥+77

f)4b^2c-(b^2+c^2-a^2)^2

=4𝑏2𝑐−1(−𝑎2+𝑏2+𝑐2)2

=4𝑏2𝑐−1(−𝑎2+𝑏2+𝑐2)(−𝑎2+𝑏2+𝑐2)

=4𝑏2𝑐−1((−𝑎2+𝑏2+𝑐2)(−𝑎2)+(−𝑎2+𝑏2+𝑐2)⋅𝑏2+(−𝑎2+𝑏2+𝑐2)⋅𝑐2)

=4𝑏2𝑐−1(−(−𝑎2+𝑏2+𝑐2)⋅𝑎2+(−𝑎2+𝑏2+𝑐2)⋅𝑏2+(−𝑎2+𝑏2+𝑐2)⋅𝑐2)

=4𝑏2𝑐−1(−(−𝑎2⋅𝑎2+𝑎2𝑏2+𝑎2𝑐2)+(−𝑎2+𝑏2+𝑐2)⋅𝑏2+(−𝑎2+𝑏2+𝑐2)⋅𝑐2)

=4𝑏2𝑐−1(−(𝑎2𝑏2+𝑎2𝑐2−𝑎2⋅𝑎2)+(−𝑎2+𝑏2+𝑐2)⋅𝑏2+(−𝑎2+𝑏2+𝑐2)⋅𝑐2)

=4𝑏2𝑐−1(−1𝑎2𝑏2−1𝑎2𝑐2−1(−𝑎2)⋅𝑎2+(−𝑎2+𝑏2+𝑐2)⋅𝑏2+(−𝑎2+𝑏2+𝑐2)⋅𝑐2)

=1(−1𝑎2𝑏2−1𝑎2𝑐2−1(−𝑎2)⋅𝑎2+(−𝑎2+𝑏2+𝑐2)⋅𝑏2+(−𝑎2+𝑏2+𝑐2)⋅𝑐2)