Question

Phân tích thành nhân tử chung:

\(\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)\)+ 1

Mong các bạn có thể mình giải bài này càng nhanh càng tốt nha!!

Answer 1

(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

Answer 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 :  \(\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)+1=\left(a^2+5a+5\right)^2\)

Answer 3

\(\left(a+1\right)\left(a+2\right)\left(a+3\right)\left(a+4\right)+1\)

\(=\left(a+1\right)\left(a+4\right)\left(a+2\right)\left(a+3\right)+1\)

\(=\left(a^2+5a+4\right)\left(a^2+5a+6\right)+1\)

\(=\left(a^2+5a+5-1\right)\left(a^2+5a+5+1\right)+1\)

\(=\left(a^2+5a+5\right)^2-1+1\)

\(=\left(a^2+5a+5\right)^2\)