\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15=0\)
\(\Leftrightarrow\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15=0\)
Đặt \(x^2+8x+11=y\Rightarrow x^2+8x+7=y-4;x^2+8x+15=y+4\)
Khi đó:
\(pt\Leftrightarrow\left(y-4\right)\left(y+4\right)+15=0\)
\(\Leftrightarrow y^2-1=0\)
\(\Leftrightarrow y=1;y=-1\)
Nếu \(y=1\Rightarrow x^2+8x+11=1\)
\(\Rightarrow x^2+8x+10=0\)
\(\Rightarrow-\left(6-x^2-8x-16\right)=0\)
\(\Rightarrow-\left[6-\left(x+4\right)^2\right]=0\)
\(\Rightarrow-\left(\sqrt{6}-x-4\right)\left(\sqrt{6}+x+4\right)=0\)
\(\Rightarrow x=-4-\sqrt{6};x=\sqrt{6}-4\)
Nếu \(y=-1\),ta có:
\(x^2+8x+11=-1\)
\(\Rightarrow x^2+8x+12=0\)
\(\Rightarrow x^2+2x+6x+12=0\)
\(\Rightarrow x\left(x+2\right)+6\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(x+6\right)=0\)
\(\Rightarrow x=-2;x=-6\)
Vậy \(x=-2;x=-6;x=-4-\sqrt{6};x=\sqrt{6}-4\)
