(a + 1)(a + 2)(a + 3)(a + 4) + 1
= (a2 + 4a + a + 4)(a2 + 3a + 2a + 6) + 1
= (a2 + 5a + 4)(a2 + 5a + 6) + 1 (1)
Đặt a2 + 5a + 5 = b
=> a2 + 5a + 4 = b - 1
a2 + 5a + 6 = b + 1
(1) = (b - 1)(b + 1) + 1
= b2 - 1 + 1
= b2
= (a2 + 5a + 5)2
\(\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)+1=\left[\left(a+1\right).\left(a+4\right)\right].\left[\left(a+2\right).\left(a+3\right)\right]+1\)
\(=\left(a^2+4a+a+4\right).\left(a^2+2a+3a+6\right)+1=\left(a^2+5a+4\right).\left(a^2+5a+6\right)+1\)
Đặt : \(a^2+5a+5=b\) thì ta có :
\(\left(b-1\right).\left(b+1\right)+1=b^2-1+1=b^2\)
thay \(a^2+5a+5\) vào b . ta được :
\(b^2=\left(a^2+5a+5\right)^2\)
VẬy : \(\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)+1=\left(a^2+5a+5\right)^2\)
\(\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)+1\)
\(=\left(a+1\right)\left(a+4\right)\left(a+2\right)\left(a+3\right)+1\)
\(=\left(a^2+5a+4\right)\left(a^2+5a+6\right)+1\)
\(=\left(a^2+5a+5-1\right)\left(a^2+5a+5+1\right)+1\)
\(=\left(a^2+5a+5\right)^2-1+1\)
\(=\left(a^2+5a+5\right)^2\)