Chọn B

B

B

A=(a+1)(a+2)(a^2+4)(a-1)(a^2+1)(a-2)

A =(a+1)(a-1)(a+2)(a-2)(a^2+4)(a^2+1)

A =(a^2-1)(a^2+1)(a^2-4)(a^2+4)

A =(a^4-1)(a^4-16)

A =\(a^{16}-16\cdot a^4-a^4+16\)

A =\(a^{16}-17\cdot a^4+16\)

B=(a+2b-3c-d)(a+2b+3c+d)

B=[(a+2b)^2 - (3c +d)^2]

B=[a^2+4ab+4b^2-(9c^2+6cd+d^2)]

B=a^3+4ab+4b^2 - 9c^2 - 6cd - d^2

C=(1-x-2x^3+3x^2)(1-x+2x^3-3x^2)

C=[(1-x)^2-(2x^3-3x^2)^2]

C=[(1-2x+x^2) - (4x^6-12x^5+9x^4)]

C=[1-2x-x^2-4x^6+12x^5-9x^4]

C=-4x^6+12x^5-9x^4-x^2-2x+1

D=(a^6-3a^3+9)(a^3+3)

D=a^9+27

a) \(\left(2x+1\right)^3\)

\(=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1+1\)

\(=8x^3+12x^2+6x+1\)

b) \(\left(x-3\right)^3\)

\(=x^3-3.x^2.3+3.x.3^2-3^3\)

\(=x^3-9x^2+27x-27\)

Bài 2: 

a: \(x^3+15x^2+75x+125=\left(x+5\right)^3\)

b: \(1-15y+75y^2-125y^3=\left(1-5y\right)^3\)

c: \(8x^3+4x^2y+\dfrac{3}{2}xy^2+8y^3=\left(2x+2y\right)^3\)

a: \(\left(2x^2+3y\right)^3\)

\(=8x^6+3\cdot4x^4\cdot3y+3\cdot2x^2\cdot9y^2+27y^3\)

\(=8x^6+36x^4y+54x^2y^2+27y^3\)

b: \(\left(2a^2b+\dfrac{1}{3}ab^2\right)^2\)

\(=4a^4b^2+2\cdot2a^2b\cdot\dfrac{1}{3}ab^2+\dfrac{1}{9}a^2b^4\)

\(=4a^4b^2+\dfrac{4}{3}a^3b^3+\dfrac{1}{9}a^2b^4\)

dấu ^ là mũ nha mn

a) \(9\left(a+b\right)^2-4\left(a-2b\right)^2\)

\(=\left[3\left(a+b\right)+2\left(a-2b\right)\right]\left[3\left(a+b\right)-2\left(a-2b\right)\right]\)

\(=\left(3a+3b+2a-4b\right)\left(3a+3b-2a+4b\right)\)

\(=\left(5a-b\right)\left(a+7b\right)\)

b) \(\left(2a-b\right)^2-4\left(a-b\right)^2\)

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

\(=\left(2a-b-2a+2b\right)\left(2a-b+2a-2b\right)\)

\(=b\left(4a-3b\right)\)

c) \(125-\left(x+2\right)^3\)

\(=\left(5-x-2\right)\left[25+5\left(x+2\right)+\left(x+2\right)^2\right]\)

\(=\left(3-x\right)\left(25+5x+10+x^2+4x+4\right)\)

\(=\left(3-x\right)\left(x^2+9x+39\right)\)

d) \(\left(x+3\right)^3-8=\left(x+3-2\right)\left[\left(x+3\right)^2+2\left(x+3\right)+4\right]\)

\(=\left(x+1\right)\left(x^2+8x+19\right)\)

e) \(x^{12}-y^4=\left(x^6\right)^2-\left(y^2\right)^2=\left(x^6-y^2\right)\left(x^6+y^2\right)\)  9 khai triển tiếp hđt 6,7)

a) Biến đổi VT . Mẫu chung là ( a + 2b )( a - 2b )

\(VT=\frac{a+2b-6b-2\left(a-2b\right)}{a^2-4b^2}=-\frac{a}{a^2-4b^2}\)( 1 )

Biến đổi VP 

\(-\frac{1}{2a}\left(\frac{a^2+4b^2}{a^2-4b^2}+1\right)=-\frac{1}{2a}\cdot\frac{a^2+4b^2+a^2-4b^2}{a^2-4b^2}\)

\(=-\frac{1}{2a}\cdot\frac{2a^2}{a^2-4b^2}=-\frac{a}{a^2-4b^2}\)( 2 )

Từ ( 1 ) và ( 2 ) => VT = VP ( đpcm )

b) \(a^3+b^3+\left(\frac{b\left(2a^3+b^3\right)}{a^3-b^3}\right)=\left(\frac{a\left(a^3+2b^3\right)}{a^3-b^3}\right)^3\)

<=> \(b^3+\left(\frac{b\left(2a^3+b^3\right)}{a^3-b^3}\right)^3=\left(\frac{a\left(a^3+2b^3\right)}{a^3-b^3}\right)-a^3\)( * )

Biến đổi VT của ( * ) ta có :

\(VT=\left[b+\frac{b\left(2a^3+b^3\right)}{a^3-b^3}\right]\left[b^2-\frac{b^2\left(2a^3+b^3\right)}{a^3-b^3}+\frac{b^2\left(2a^3+b^3\right)^2}{\left(a^3-b^3\right)^2}\right]\)

\(=\frac{3a^3b}{a^3-b^3}\cdot\frac{3a^6b^2+3a^3b^5+3b^8}{\left(a^3-b^3\right)^2}\)

\(=\frac{9a^3b^3}{\left(a^3-b^3\right)^3}\left(a^6+a^3b^3+b^6\right)\)( 1 )

\(VP=\left[\frac{a\left(a^3+2b^3\right)}{a^3-b^3}-a\right]\left[\frac{a^2\left(a^3+2b^3\right)^2}{\left(a^3-b^3\right)^2}+\frac{a^2\left(a^3+2b^3\right)}{a^3-b^3}+a^2\right]\)

\(=\frac{3ab^3}{a^3-b^3}\cdot\frac{3a^8+3a^5b^3+3a^2b^6}{\left(a^3-b^3\right)^2}\)

\(=\frac{9a^3b^3}{\left(a^3-b^3\right)^3}\left(a^6+a^3b^3+b^6\right)\)( 2 )

Từ ( 1 ) và ( 2 ) => VT = VP => ( * ) đúng 

=> Hằng đẳng thức đúng 

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 !