\(8x^3+12x^2y+6xy^2+y^3-z^3\)

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

\(=\left(2x+y-z\right)\left[4x^2+z\left(2x+y\right)+z^2\right]\)

a, 8a3 - 36a2 +54ab2 - 27b3

=(8a3-36a2b +54ab2 - 27b3)

=(2a-3b)2

=(2a-3b)(2a-3b)(2a-3b)

b, 8x3 + 12x2y + 6xy2 + y3 - z 3

=(8x3 + 12x2y + 6xy2 + y3) - z3

=(2x + y)3 - y3

=(2x + y +z) . [ (2x + Y)2 + 2(2x + y)+ z2

= (2x + y + z)(4x2 + 4xy + y2 + 4x + 2y + z2

a, 4\(x^3\).y + \(\dfrac{1}{2}\)yz

  =y.(4\(x^3\) + \(\dfrac{1}{2}\)z)

b, (a² + b² - 5)² - 2.(ab + 2)²

 = [a² + b² - 5  - \(\sqrt{2}\)(ab + 2) ].[ a² + b² - 5 + \(\sqrt{2}\)(ab +2)]

a) \(4x^3y+\dfrac{1}{2}yz=y\left(4x^3+\dfrac{1}{2}z\right)\)

b) \(\left(a^2+b^2-5\right)^2-2.\left(ab+2\right)^2\)

\(=\left[\left(a^2+b^2-5\right)+2\left(ab+2\right)\right]\left[\left(a^2+b^2-5\right)-2\left(ab+2\right)\right]\)

\(=\left[a^2+b^2-5+2ab+4\right]\left[a^2+b^2-5-2ab-4\right]\)

\(=\left[a^2+b^2+2ab-1\right]\left[a^2+b^2-2ab-9\right]\)

\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-9\right]\)

\(=\left[\left(a+b+1\right)\left(a+b-1\right)\right]\left[\left(a-b+3\right)\left(a-b-3\right)\right]\)

Mình đã làm sai, cô Hoài Thương đúng rồi.

b, A=[(a+1)(a+7)][(a+3)(a+5)]+15

=>A=(a²+8a+7)(a²+8a+15)+15

Đặt a²+8a+11= t

=>a²+8a+7= t-4 và a²+8a+15= t+4

=>A=(t-4)(t+4)+15

=>A=t²-16+15

      =t²-1=(t-1)(t+1)

Thay t = a²+8a+11

=>A=(a²+8a+11-1)(a²+8a+11+1)

=>A=(a²+8a+10)(a²+8a+12)

a)   \(x^2+2xy+7x+7y+y^2+10\)

\(=\left(x+y\right)^2+7\left(x+y\right)+\frac{49}{4}-\frac{9}{4}\)

\(=\left(x+y+\frac{7}{2}\right)^2-\frac{9}{4}\)

\(=\left(x+y+\frac{7}{2}-\frac{3}{2}\right)\left(x+y+\frac{7}{2}+\frac{3}{2}\right)\)

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

c,=(X⁵ - X²)+(X² +X+1)

  =X²(X³-1)+(X²+X+1)

  =X²(X-1)(X²+X+1)+(X²+X+1)

=(X³-X²+1)(X²+X+1)

\(A=\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

\(=\left[\left(a+1\right)\left(a+7\right)\right]\left[\left(a+3\right)\left(a+5\right)\right]+15\)

\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)

Đặt: \(a^2+8a+11=t\), khi đó pt trở thành:

\(\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1=\left(t-1\right)\left(t+1\right)\)

\(=\left(a^2+8a+11-1\right)\left(a^2+8a+11+1\right)=\left(a^2+8a+10\right)\left(a^2+8a+12\right)\\ =\left(a+2\right)\left(a+6\right)\left(a^2+8a+10\right)\)

\(A=\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)

Đặt \(t=a^2+8a+7\) khi đó A thành:

\(t\left(t+8\right)+15=t^2+8t+15\)

\(=\left(t+3\right)\left(t+5\right)=\left(a^2+8a+7+3\right)\left(a^2+8a+7+5\right)\)

\(=\left(a^2+8a+10\right)\left(a^2+8a+12\right)\)

\(=\left(a^2+8a+10\right)\left(a+2\right)\left(a+6\right)\)

Ta có:

\(\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

\(\left[\left(a+1\right)\left(a+7\right)\left(a+3\right)\left(a+5\right)\right]+15\)

\(\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)

Đặt \(a^2+8a+7=t\)

\(\Rightarrow t\left(t+8\right)+15\)

\(=t^2+8t+15\)

\(=t\left(t+3\right)+5\left(t+3\right)\)

\(=\left(t+3\right)\left(t+5\right)\)

\(\Rightarrow\left[\left(a^2+8a+7\right)+3\right]\left[\left(a^2+8a+7\right)+5\right]\)

\(=\left(a^2+8a+10\right)\left(a^2+8a+12\right)\)

\(=\left(a^2+8a+10\right)\left(a^2+2a+6a+12\right)\)

\(=\left(a^2+8a+10\right)\left[a\left(a+2\right)+6\left(a+2\right)\right]\)

\(=\left(a^2+8a+10\right)\left(a+2\right)\left(a+6\right)\)

\(A=\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

\(=\left[\left(a+1\right)\left(a+7\right)\right]\left[\left(a+3\right)\left(a+5\right)\right]+15\)

\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)

Đặt : \(a^2+8+11=t\) khi đó pt trở thành :

\(\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1=\left(t-1\right)\left(t+1\right)\)

\(=\left(a^2+8a+11-1\right)\left(a^2+8a+11+1\right)\)

\(=\left(a^2+8a+10\right)\left(a^2+8a+12\right)\)

\(=\left(a+2\right)\left(a+6\right)\left(a^2+8a+10\right)\)

Chúc bạn học tốt !!!

A = (a+1)(a+3)(a+5)(a+7) + 15

A = [ (a+1) (a+7)] [(a+3) (a+5)] + 15

A= ( a² + 8a + 7)( a² + 8a + 15 ) + 15                 (*)

         Đặt a² + 8a + 7 = t

=> A = t.(t+8) + 15

     A = t² + 8t + 15

     A = t² + 3t + 5t + 15

     A = ( t +3).(t+5)

  Thay   A = ( t +3).(t+5) vào (*)

=> A = ( a² + 8a + 7 + 3).( a² + 8a + 7 + 5)

    A = ( a² + 8a + 10).( a² + 8a + 12 )

     A = ( a² + 8a + 10).( a² + 6a + 2a + 12 )

       A = ( a² + 8a + 10) ( a+6)(a+2)

a=(a+1)(a+3)(a+5)(a+7)+15

a=[ (a+1)(a+7) ] [(a+3)(a+5)] +15

a=(a²+8a+7)(a²+8a+15)+15

        Đặt a²+8a+7=t

         a=t.(t+8)+15

         a=t²+8t+15

          a=t²+3t+5t+15

         a=(t+3).(t+5)

Hok tok

Đặt \(M=\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)

\(M=\left[\left(a+1\right)\left(a+7\right)\right]\left[\left(a+3\right)\left(a+5\right)\right]+15\)

\(M=\left(a^2+7a+a+7\right)\left(a^2+5a+3a+15\right)+15\)

\(M=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)

Đặt \(p=a^2+8a+11\)

\(\Rightarrow M=\left(p-4\right)\left(p+4\right)+15\)

\(\Rightarrow M=p^2-16+15\)

\(\Rightarrow M=p^2-1\)

\(\Rightarrow M=\left(p-1\right)\left(p+1\right)\)

Thay \(p=a^2+8a+11\)vào M, ta có :

\(M=\left(a^2+8a+11-1\right)\left(a^2+8a+11+1\right)\)

\(M=\left(a^2+8a+10\right)\left(a^2+8a+12\right)\)

(a+1)(a+7)(a+3)(a+5)+15
=(a²+8a+7)(a²+8a+15)+15
=(a²+8a+11-4)(a²+8a+11+4)+15
=(a²+8a+11)²-4²+15
=(a²+8a+11)²-1
=(a²+8a+11-1)(a²+8a+11+1)
=(a²+8a+10)(a²+8a+12)
 

Rút gọn biểu thức sau:A=(2x-3)(2x+3)-(x+5)²-(x-1)(x+2)

(a+b)³-(a-b)³=a³+3a²b+3ab²+b³-(a³-3a²b+3ab²-b³)

                    =a³+3a²b+3ab²+b³-a³+3a²b-3ab²+b³

                        =6a²b+2b³

Áp dụng hđt a³-b³=(a-b)(a²+ab+b²) ấy

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

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

\(=2b\left(3a^2+b^2\right)\)

Ta có : \(\left(a+b\right)^3-\left(a-b\right)^3\) 

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

\(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3\)

\(=6a^2b+2b^3\)

A=( a +1)(a+3)(a+5)(a+7)+15

=(a+1)(a+7)(a+3)(a+5)+15

=(a²+8a+7)(a²+8a+15)+15

Đặt y=a²+8a+7 ta được :

y(y+8)+15=y² + 8y +15

=y² +3y+5y+15

=y(y+3) +5(y+3)

=(y+3)(y+5)

thay y=a²+8a+7 ta được 

(a²+8a+7+3)(a²+8a+7+5)

=(a²+8a+10)(a²-2a-6a+12)

=(a²+8a+10)[a(a-2)-6(a-2)]

=(a²+8a+10)(a-2)(a-6)