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

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

\(A=\left(a+1\right).\left(a+3\right).\left(a+17\right)+15\)

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

\(=\left(a^2+8a+11-4\right)\left(a^2+8a+11+4\right)+15\)

\(=\left(a^2+8a+11\right)^2-4^2+15\)

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

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

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

(a²+8a+12) = (a²+2a+6a+12)

                   = (a+2) (a+6)

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

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

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

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

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)

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

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

A=(a²+8a+7)(a²+8a+15)+15

Đặt a²+8a = v

Ta có : 

A=(v+7)(v+15)+15

A= v²+22v+105+15

A= v²+22v+ 120

A= v²+10v+12v+120

A=( v²+10v)+(12v+120)

A=[v(v+10)]+[12(v+10)]

A=(v+10)(v+12)             (1)

Thay a²+8a = v vào (1) 

A=(a²+8a+10)(a²+8a+12) 

=(a^2+8a+7)*(a^2+8a+15)+15

Đặt (a^2+8a+7)=t ta có

t*(t+8)+15=t^2+8t+15=t^2+3t+5t+15=(t+3)*(t+5)(*)

Thay t=a^2+8a+7 vào (*) là được

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

=(a^2+8a+7)(a^2+8a+15)+15

=(a^2+8a+11-4)(a^2+8a+11+4)+15

=(a^2+8a+11)^2 -4^2+15

=(a^2+8a+11)^2 -1

=(a^2+8a+11-1)(a^2+8a+11+1)

=(a^2+8a+10)(a^2+8a+12)

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

=(a^2+8a+7)(a^2+8a+15)+15

=(a^2+8a+11-4)(a^2+8a+11+4)+15

=(a^2+8a+11)^2 -4^2+15

=(a^2+8a+11)^2 -1

=(a^2+8a+11-1)(a^2+8a+11+1)

=(a^2+8a+10)(a^2+8a+12)

1)(x^2+3x+1)(x^2+3x+2)-6

Đặt t = x²  + 3x + 1

Khi đó PT có dạng:

t.(t + 1) - 6

= t² + t - 6

= t² - 2t - 3t - 6

= t.(t - 2) + 3.(t - 2)

= (t + 3).(t - 2)

= (x² + 3x + 1 + 3).(x² + 3x + 1 - 2)

= (x² + 3x + 4).(x² + 3x - 1)

\(1\hept{\begin{cases}\left(x^2+3x+2-1\right)\left(x^2+2x+2\right)-6\\\left(t-1\right)\left(t\right)-6\\t^2-t-6\end{cases}}.\) " đặt x^2+3x+2 = t

\(\hept{\begin{cases}t^2-\frac{2t.1}{2}+\frac{1}{4}-\left(\frac{24+1}{4}\right)\\\left(t-\frac{1}{2}\right)^2-\frac{25}{4}\\\left(t-\frac{1}{2}\right)^2-\frac{25}{4}\end{cases}}\)

\(\hept{\begin{cases}\left(t-\frac{1}{2}-\frac{5}{2}\right)\left(t-\frac{1}{2}+\frac{5}{2}\right)\\\left(t-\frac{7}{2}\right)\left(t+\frac{4}{2}\right)\\\left(t-\frac{7}{2}\right)\left(t+\frac{4}{2}\right)\end{cases}}\)

2)  \(\hept{\begin{cases}\left\{\left(x+1\right)\left(x+7\right)\right\}\left\{\left(x+5\right)\left(x+3\right)\right\}+15\\\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\\t\left(t+8\right)+15\end{cases}}\)  

\(\hept{\begin{cases}t^2+8t+15\\\left(t^2+8t+16\right)-1\\\left(t+4\right)^2-1\end{cases}}\Leftrightarrow\left(t+5\right)\left(t+4\right)\)

\(\hept{\begin{cases}a^3\left(b-c\right)+b^3\left(c-a+b-b\right)+c^3\left(a-b\right)\\a^3\left(b-c\right)-b^3\left(-c+a-b+b\right)+c^3\left(a-b\right)\\a^3\left(b-c\right)-b^3\left(a-b\right)-b^3\left(b-c\right)+c^3\left(a-b\right)\end{cases}\Leftrightarrow\hept{\begin{cases}\left(b-c\right)\left(a^3-b^3\right)-\left(a-b\right)\left(b^3-c^3\right)\\\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)-\left(a-b\right)\left(b-c\right)\left(b^2+ab+c^2\right)\\\left(a-b\right)\left(b-c\right)\left(a^2+2ab+2b^2+c^2\right)\end{cases}}}\)