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Lời giải:
$-4x^4+1+2x^2y-y=-(4x^4-1)+(2x^2y-y)$
$=-(2x^2-1)(2x^2+1)+y(2x^2-1)$
$=(2x^2-1)[-(2x^2+1)+y]$
$=(2x^2-1)(y-2x^2-1)$
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\(x^4+4x^3+2x^2-4x+1\)
\(=x^4+2x^3-x^2+2x^3+4x^2-2x-x^2-2x+1\)
\(=x^2\left(x^2+2x-1\right)+2x\left(x^2+2x-1\right)-\left(x^2+2x-1\right)\)
\(=\left(x^2+2x-1\right)^2\)
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\(4x^4-8x^3+4x^3-8x^2+x^2-2x-2x+4\\ =4x^3\left(x-2\right)+4x^2\left(x-2\right)+x\left(x-2\right)-2\left(x-2\right)\\ =\left(x-2\right)\left(4x^3+4x^2+x-2\right)\\ =\left(x-2\right)\left(4x^3-2x^2+6x^2-3x+4x-2\right)\\ =\left(x-2\right)\left[2x^2\left(2x-1\right)+3x\left(2x-1\right)+2\left(2x-1\right)\right]\\ =\left(x-2\right)\left(2x-1\right)\left(2x^2+3x-2\right)\)
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x4 - 4x - 1 = x4 -2x2 + 1 - 2x2 +4x - 2
= (x2 - 1)2 - 2(x - 1)2
= ... Tự làm tiếp nhé
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\(=4x^4+4x^2+1-4x^2=\left(2x^2+1\right)^2-4x^2=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)\)