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

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

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

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 \(a^2+8a+7=t\)

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

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

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

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

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

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

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

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

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

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+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)

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

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

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

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

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

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

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

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

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

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

(a²+8a+12) = (a²+2a+6a+12)

                   = (a+2) (a+6)

a)x^5+x+1

=x⁵-x²+x²+x+1

=x²(x³-1)+x²+x+1

=x²(x+1)(x²+x+1)+x²+x+1

=(x²+x+1)(x³+x²+1)

b)(x+1)(x+3)(x+5)(x+7)+15

=(x²+8x+7)(x²+8x+15)+15

Đặt x²+8x+7=t

=> t(t+8)+15=t²+8t+15

=(t+3)(t+5)

=(x²+8x+10)(x²+8x+12)