Question

[Hikaru Akira]

Viết các đa thức sau thành tích                                                                            

1 ) (3x + 2)² – 4

2) 4x² – 25y²

3 ) 4x² – 49                                                      

4) 8z³ + 27

5, \(\dfrac{9}{25}\)x⁴  – \(\dfrac{1}{4}\)                                                    

6, x³² – 1

Mình cần quá trình làm nữa nha, thanks 

Answer 1

[ILoveMath]

1) \(\left(3x+2\right)^2-4\\ =\left(3x+2\right)^2-2^2\\ =\left(3x+2-2\right)\left(3x+2+2\right)\\ =3x.\left(3x+4\right)\)

2) \(4x^2-25y^2=\left(2x\right)^2-\left(5y\right)^2=\left(2x-5y\right)\left(2x+5y\right)\)

3) \(4x^2-49=\left(2x\right)^2-7^2=\left(2x-7\right)\left(2x+7\right)\)

4) \(8z^3+27=\left(2z\right)^3+3^3=\left(2z+3\right)\left(4z^2+6z+9\right)\)

5) \(\dfrac{9}{25}x^4-\dfrac{1}{4}=\left(\dfrac{3}{5}x^2\right)^2-\left(\dfrac{1}{2}\right)^2=\left(\dfrac{3}{5}x^2-\dfrac{1}{2}\right)\left(\dfrac{3}{5}x^2+\dfrac{1}{2}\right)\)

6) \(x^{32}-1\\ =\left(x^{16}\right)^2-1^2\\ =\left(x^{16}-1\right)\left(x^{16}+1\right)\\ =\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\\ =\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\\ =\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)

Answer 2

[Nguyễn Lê Phước Thịnh]

1: \(\left(3x+2\right)^2-4=3x\left(3x+4\right)\)

2: \(4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)

3: \(4x^2-49=\left(2x-7\right)\left(2x+7\right)\)

4: \(8z^3+27=\left(2z+3\right)\left(4z^2-6z+9\right)\)

5: \(\dfrac{9}{25}x^4-\dfrac{1}{4}=\left(\dfrac{3}{5}x^2-\dfrac{1}{2}\right)\left(\dfrac{3}{5}x^2+\dfrac{1}{2}\right)\)