Bài 1:
a. $x^3+6x^2+11x+6=x^3+1+6x^2-6+11x+11$
$=(x+1)(x^2-x+1)+6(x+1)(x-1)+11(x+1)$
$=(x+1)(x^2-x+1+6x-6+11)=(x+1)(x^2+5x+6)$
$=(x+1)(x+2)(x+3)$
b. $P(x)=x^4-8x^2-x+12$
$=x^4-7x^2+\frac{49}{4}-x^2-x-\frac{1}{2}$
$=(x^2-\frac{7}{2})^2-(x+\frac{1}{2})^2$
$=(x^2+x-3)(x^2-x-4)$
$=(x-\frac{-1+\sqrt[]{13}}{2})(x-\frac{-1-\sqrt[]{13}}{2})(x-\frac{1+\sqrt[]{17}}{2})(x-\frac{1+\sqrt[]{17}}{2})$
Bài 2:
Xét $2(x^2+y^2)-(x+y)^2=2x^2+2y^2-(x^2+2xy+y^2)$
$=x^2-2xy+y^2=(x-y)^2≥0$
Vậy $ 2(x^2+y^2)-(x+y)^2≥0$
⇔ $x^2+y^2≥$$\frac{(x+y)^2}{2}$
⇔ $\frac{x^2+y^2}{2}≥$$\frac{(x+y)^2}{2}$
Dấu "=" xảy ra khi $x-y=0$ ⇔ $x=y$
