Bài 1:
a) Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)(*)
Đặt \(a=x^2+5x\)
(*)\(=\left(a+4\right)\left(a+6\right)+1\)
\(=a^2+10a+24+1\)
\(=a^2+10a+25\)
\(=\left(a+5\right)^2\)
\(=\left(x^2+5x+5\right)^2\)
b) Ta có: \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)(**)
Đặt \(b=x^2+8x\)
(**)\(=\left(b+7\right)\left(b+15\right)+15\)
\(=b^2+22b+105+15\)
\(=b^2+22b+120\)
\(=b^2+12b+10b+120\)
\(=b\left(b+12\right)+10\left(b+12\right)\)
\(=\left(b+12\right)\left(b+10\right)\)
\(=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)
\(=\left(x+2\right)\left(x+6\right)\left(x^2+8x+10\right)\)
c) Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)(***)
Đặt \(c=x^2+5x\)
(***)\(=\left(c+4\right)\left(c+6\right)-24\)
\(=c^2+10c+24-24\)
\(=c^2+10c\)
\(=c\left(c+10\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
\(=x\left(x+5\right)\left(x^2+5x+10\right)\)
