Ta có:
\(\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
Đặt x2 + 10x + 16= t thì:
\(t\left(t+8\right)+16=t^2+8t+16\)
\(=t^2+4t+4t+16=\left(t+4\right)^2\)
\(=\left(x^2+10+20\right)^2\)
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