(x+1)(x+2)(x+3)(x+4)-8
=[(x+1).(x+4)].[(x+2).(x+3)]-8
=(x2+5x+4).(x2+5x+6)-8
Đặt (x2+5x+4)=t =>(x2+5x+6)=t+2
Thay vào biểu thức ta có:
(x2+5x+4).(x2+5x+6)-8
t.(t+2)-8
=t2+2t+1-9
=(t+1)2-32
=(x2+5x+4+1)-32
=(x2+5x+5+3).(x2+5x+5-3)
=(x2+5x+8).(x2+5x+2)
=
ta làm như sau :
\(\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-8.\)
\(\Rightarrow\left(x^2+5X+4\right)\left(x^2+5x+6\right)-8\)
Đặt \(x^2+5x+4=t\)
\(\Leftrightarrow t\left(t+2\right)-8\)
\(\Leftrightarrow t^2+2t-8\Leftrightarrow t^2+2t+1-9\)
\(\Leftrightarrow\left(t+1\right)^2-3^2\)
\(\Leftrightarrow\left(t-2\right)\left(t+4\right)\)
\(\Leftrightarrow\left(x^2+5x+2\right)\left(x^2+5x+8\right)\)
