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S

su10112000a

$x(x+1)(x+2)(x+3)+1$
$=x(x+3)(x+1)(x+2)+1$
$=(x^2+3x)(x^2+3x+2)+1$
đặt $x^2+3x+1=y$, ta có:
$(y-1)(y+1)+1$
$=y^2-1+1$
$=y^2$
Thay $y=x^2+3x+1$, ta có:
$(x^2+3x+1)^2$
$=[(x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4})-\dfrac{5}{4}]^2$
$=[(x+\dfrac{3}{2})^2-\dfrac{5}{4}]^2$
$=(x+\dfrac{3}{2}-\dfrac{\sqrt{5}}{2})^2(x+\dfrac{3}{2}+\dfrac{\sqrt{5}}{2})^2$
 
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