5:

a: (2x-5)(2x+5)=4x^2-25

b: (3x-5y)(3x+5y)=9x^2-25y^2

c: (3x+7y)(3x-7y)=9x^2-49y^2

d: (2x-1)(2x+1)=4x^2-1

4:

a: 2003*2005=(2004-1)(2004+1)=2004^2-1<2004^2

b: 8(7^2+1)(7^4+1)(7^8+1)

=1/6*(7-1)(7+1)(7^2+1)(7^4+1)(7^8+1)

=1/6(7^2-1)(7^2+1)(7^4+1)(7^8+1)

=1/6(7^16-1)<7^16-1

5:

a: (2x-5)(2x+5)=4x^2-25

b: (3x-5y)(3x+5y)=9x^2-25y^2

c: (3x+7y)(3x-7y)=9x^2-49y^2

d: (2x-1)(2x+1)=4x^2-1

mik chỉ biết bài 5 thôi !

\(1,=x\left(2x+3\right)-\left(2x+3\right)=2x^2+3x-2x-3=2x^2+x-3\\ 2,=\left(2x\right)^2-2.2.x.1+1^2=4x^2-4x+1\)

1. =2x²-2x+3x-3

2. (2x)²-2.2x.1+1²

=4x²-4x+1

\(4x^2-12x+9\)

(2x+3)²

a) \(=4x^2-12x+9\)

b) \(=4x^2+2x+\dfrac{1}{4}\)

c) \(=4x^2-\dfrac{4}{3}x+\dfrac{1}{9}\)

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

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

f) \(=\left(125-4x\right)\left(125^2+500x+16x^2\right)\)

1) \(\left(x+1\right)^2=x^2+2x+1\)

2) \(\left(2x+1\right)^2=4x^2+4x+1\)

3) \(\left(2x+y\right)^2=4x^2+4xy+y^2\)

4) \(\left(2x+3\right)^2=4x^2+12x+9\)

5) \(\left(3x+2y\right)^2=9x^2+12xy+4y^2\)

6) \(\left(2x^2+1\right)^2=4x^4+4x^2+1\)

7) \(\left(x^3+1\right)^2=x^6+2x^3+1\)

8) \(\left(x^2+y^3\right)^2=x^4+2x^2y^3+y^6\)

9) \(\left(x^2+2y^2\right)^2=x^4+4x^2y^2+4y^4\)

10) \(\left(\dfrac{1}{2}x+\dfrac{1}{3}y\right)^2=\dfrac{1}{4}x^2+\dfrac{1}{3}xy+\dfrac{1}{9}y^2\)

a)
x³+3x²+3x+1
b)8x³+18x²+54x+27
c)x³+3636x²+3636x+1780360128
d)x⁶-6x²+12x-8
e)
8x³-36x²y+54xy²-27y³

\(a,=x^3+3x^2+3x+1\\ b,=8x^3+36x^2+54x+27\\ c,=x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}\\ d,=x^6-6x^4+12x^2-8\\ e,=8x^3-36x^2y+54xy^2-27y^3\)

Bài 1:

a, \(\left(x-y+2z\right)^2=x^2+y^2+4z^2-2xy-4yz+4zx\)

b, \(\left(2x-3\right)\left(2x+3\right)\left(4x^2+9\right)=\left(4x^2-9\right)\left(4x^2+9\right)=16x^4-81\)