Có \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+2008\)
\(=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+2008\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+2008\)
Đặt \(x^2+10x+21=a\)
\(\Rightarrow\) \(\left(a-5\right)\left(a+3\right)+2008\)
\(=a^2-2a-15+2008\)
\(=a\left(a-2\right)+1993\)
\(=\left(x^2+10x+21\right)\left(x^2+10x+19\right)+1993\)
Có \(\left(x^2+10x+21\right)\left(x^2+10x+19\right)⋮\left(x^2+10x+21\right)\) \(\)
\(\Rightarrow\)\(\left(x^2+10x+21\right)\left(x^2+10x+19\right)+1993\div\left(x^2+10x+21\right)\) dư 1993
Ta có: \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+2008\)
\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+2008\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+2008\)(*)
Đặt \(a=x^2+10x+21\)
\(\Rightarrow x^2+10x+16=a-5\)
\(x^2+10x+24=a+3\)
Thay \(x^2+10x+16=a-5;x^2+10x+24=a+3\) vào (*) ta được:
\(\left(a-5\right)\left(a+3\right)+2008\)
\(=a^2-2a-15+2008\)
\(=a\left(a-2\right)+1993\)
Vì \(a\left(a-2\right)⋮1993\Rightarrow a\left(a-2\right)+1993\) chia a dư 1993
hay \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+2008\) chia \(x^2+10x+21\) dư 1993
