Lời giải:
a.
$x^8+x^4+1=(x^4)^2+2x^4+1-x^4$
$=(x^4+1)^2-(x^2)^2=(x^4+1-x^2)(x^4+1+x^2)$
$=(x^4+1-x^2)[(x^2+1)^2-x^2]$
$=(x^4-x^2+1)(x^2+1-x)(x^2+1+x)$
b.
$x^{12}-3x^6-1=(x^6-\frac{3}{2})^2-\frac{13}{4}$
$=(x^6-\frac{3}{2}-\frac{\sqrt{13}}{2})(x^6-\frac{3}{2}+\frac{\sqrt{13}}{2})$
c.
$3x^4+10x^2-25=(3x^4+15x^2)-(5x^2+25)$
$=3x^2(x^2+5)-5(x^2+5)=(x^2+5)(3x^2-5)$
$=(x^2+5)(\sqrt{3}x-\sqrt{5})(\sqrt{3}x+\sqrt{5})$
c.
$x^2-5y^2-y^4+2xy-9$
$=(x^2+2xy+y^2)-(y^4+6y^2+9)$
$=(x+y)^2-(y^2+3)^2$
$=(x+y+y^2+3)(x+y-y^2-3)$
\(a,x^8+x^4+1\\ =\left(x^8+2x^4+1\right)-x^4\\ =\left(x^4+1\right)^2-x^4\\ =\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\\ b,x^{12}-3x^6-1\\ =\left(x^{12}-2x^6+1\right)-x^6-2\\ =\left(x^6-1\right)^2-x^6-2\\ =\left(x^6-x^3-1\right)\left(x^6+x^3-1\right)-2???\\ c,3x^4+10x^2-25\\ =4x^4-\left(x^4-10x^2+25\right)\\ =4x^4-\left(x^2-5\right)^2\\ =\left(2x^2-x^2+5\right)\left(2x^2+x^2-5\right)\\ =\left(x^2+5\right)\left(3x^2-5\right)\\ d,x^2-5y^2-y^4+2xy-9\\ =\left(x^2+2xy+y^2\right)-\left(y^4+6y^2+9\right)\\ =\left(x+y\right)^2-\left(y^2+3\right)^2\\ =\left(x+y+y^2+3\right)\left(x+y-y^2-3\right)\)
a) x8+x4+1 = (x4+1)2-x4 = (x4-x2+1)(x4+x2+1)
b) x12-3x6-1 = (x6-1)2-x6 = (x6-x3-1)(x6+x3-1)
c) 3x4+10x2-25 = 4x4-(x4-10x2+25) = 4x4- (x2-5)2 = (x2+5)(3x2-5)
d) x2-5y2-y4+2xy-9 = (x+y)2-(y2+3)2 = (x+y-y2-3)(x+y+y2+3)
c: Ta có: \(3x^4+10x^2-25\)
\(=3x^4+15x^2-5x^2-25\)
\(=3x^2\left(x^2+5\right)-5\left(x^2+5\right)\)
\(=\left(x^2+5\right)\left(3x^2-5\right)\)
