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 !

\(a,\left(x+2\right)^2=x^2+4x+4\\ b,\left(x-1\right)^2=x^2-2x+1\\ c,\left(x^2+y^2\right)^2=x^4+2x^2y^2+y^4\)

a) = x² + 4x + 4

b) = x² - 2x + 1

c) x⁴ + 2x²y² + y⁴

Ta có:(a²+ab+b²)(a²-ab+b²)-(a⁴+b⁴)

    =   (a²+b²)²-a²b²-a⁴-b⁴=a⁴+2a²b²+b⁴-a²b²-a⁴-b⁴=a²b²

Ta có:(a²+ab+b²)(a²-ab+b²)-(a⁴+b⁴)

    =   (a²+b²)²-a²b²-a⁴-b⁴=a⁴+2a²b²+b⁴-a²b²-a⁴-b⁴=a²b²

Chọn B

B

(a+b+c)(a+b-c)

=[(a+b) + c] [ (a+b) - c]

= (a+b)² - c²

Áp dụng HĐT: (A-B)(A+B) = A² - B²

(a+b+c).(a+b-c)

=( a+b )^2  - c^2

= a^2 + 2ab + b^2 - c^2

\(a,=x^2+4x+4\\ b,=x^3+3x^2+3x+1\\ c,=\left(x-3\right)\left(x+3\right)\)

a,\(\left(x+2\right)^2=x^2+2.x.2+2^2=x^2+4x+4\)

b, \(\left(x+1\right)^3=x^3+3.x^2.1+3.x.1^2+1^3=x^3+3x^2+3x+1\)

c,\(x^2-3^2=\left(x-3\right).\left(x+3\right)\)

a) \((a+y)^3=a^3+3a^2y+3ay^2+y^3\)

b) \((x-b)^3=x^3-3x^2b+3xb^2-b^3\)

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

b) \(\left(5p-q\right)^2=25p^2-10pq+q^2\)

c) \(\left(-a-b\right)^2=-a^2-2ab-b^2\)

d) \(\left(1+3s\right)^2=1+6s+9s^2\)

e) \(\left(a^2b+2b\right)^2=a^4b^2+4a^2b^2+4b^2\)

f) \(\left(3u-v\right)^3=27u^3-27u^2v+9uv^2-v^3\)

a,\(\left(2x-3y\right)=\left(2x\right)^2-2.2x.3y+\left(3y\right)^2\)

=\(4x^2-12xy+6y^2\)

b,\(\left(5p-q\right)^2=\left(5p\right)^2-2.5p.q+q^2\)

=\(25p^2-10pq+q^2\)

c,(-a-b)\(^2=\left(-a\right)^2-2.\left(-a\right).b+b^2\)

=\(a^2+2ab+b^2\)

d,\(\left(1+3s\right)^2=1+6s+9s^2\)

e,(a\(^2b+2b)^2=(a^2b)^2+2.a^2b.2b^2+\left(2b\right)^2\)

=\(a^4b^2+4a^2b^2+4b^2\)

f,\(\left(3u-v\right)^3=27u^3-27u^2v+9uv^2-v^3\)

1) \(\left[\left(a+b\right)-c\right]^2=\left(a+b\right)^2-2c\left(a+b\right)+c^2\)

\(=\left(a^2+2ab+b^2\right)-2ac-2bc+c^2\)

\(=a^2+b^2+c^2+2ab-2ac-2bc\)

2)Phần này tg tự

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