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\(a,=\dfrac{1}{2}\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\\ =\dfrac{1}{2}\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\dfrac{1}{2}\left(x-y\right)^2\left(x+y\right)^2\\ b,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ c,=\dfrac{1}{2}\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)^2\)
\(x-\sqrt{x}-6=\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)\)
\(2x+5\sqrt{x}-3=\left(\sqrt{x}+3\right)\left(2\sqrt{x}-1\right)\)
a) x4 + 2x3 + x2
= x2 ( x2 + 2x + 1 )
= x2 ( x + 1 )2
b) 5x2 - 10xy + 5y2 - 20z2
= 5 [(x2 - 2xy + y2 ) - 4z2 ]
= 5 [( x - y )2 - ( 2z )2 ]
= 5 ( x - y - 2z ) ( x - y + 2z )
c) x3 - x + 3x2y + 3xy2+ y3- y
= ( x3 + 3x2y + 3xy2 + y3 ) - ( x + y )
= (x + y )3 - ( x + y)
= ( x + y ) [( x + y )2 - 1 ]
= ( x + y ) ( x + y + 1 ) ( x + y - 1 )
\(x^3+2x-3\)
\(=x^3-x^2+x^2-x+3x-3\)
\(=x^2\left(x-1\right)+x\left(x-1\right)+3\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+3\right)\)
\(C=x^2\left(x^2+x+1\right)-2x\left(x^2+x+1\right)+3\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-2x+3\right)\)