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Lời giải:
$x^3-4x^2-12x+27$
$=(x^3+3x^2)-(7x^2+21x)+(9x+27)$
$=x^2(x+3)-7x(x+3)+9(x+3)$
$=(x+3)(x^2-7x+9)$

\(x^4+4x^3-2x^2-12x+9\)
\(=x^4+3x^3+x^3+3x^2-5x^2-15x+3x+9\)
\(=x^3\left(x+3\right)+x^2\left(x+3\right)-5x\left(x+3\right)+3\left(x+3\right)\)
\(=\left(x+3\right)\left(x^3+x^2-5x+3\right)\)
\(=\left(x+3\right)\left(x^3+3x^2-2x^2-6x+x+3\right)\)
\(=\left(x+3\right)\left[x^2\left(x+3\right)-2x\left(x+3\right)+\left(x+3\right)\right]\)
\(=\left(x+3\right)\left(x+3\right)\left(x^2-2x+1\right)\)
\(=\left(x+3\right)^2\left(x-1\right)^2\)

\(x^3-4x^2+12x-27\)
\(=x^3-3x^2-x^2+3x+9x-27\)
\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)

a) x2 - 4 + (x - 2)2 = 0
=> (x - 2)(x + 2) + (x2 - 4x + 4) = 0
x2 + 2x - 2x - 4 + x2 - 4x + 4 = 0
2x2 - 4x = 0
2x(x - 2) = 0
=> 2x = 0 => x = 0
x - 2 = 0 x = 2




12x2 - 12x +3
= 3(4x2- 4x +1)
= 3(2x +1)2