phân tích đa thức thành nhân tử:
x4 + (x - 1)(x2 - 2x + 2)
help me!!!
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x4−2x3+2x−1x4−2x3+2x−1
=x4−x3−x3+x2−x2+x+x−1=x4−x3−x3+x2−x2+x+x−1
=x3(x−1)−x2(x−1)−x(x−1)+(x−1)=x3(x−1)−x2(x−1)−x(x−1)+(x−1)
=(x−1)(x3−x2−x+1)=(x−1)(x3−x2−x+1)
=(x−1)[
\(x^4+2x^3+2x^2+2x+1\\ =\left(x^4+x^3\right)+\left(x^3+x^2\right)+\left(x^2+x\right)+\left(x+1\right)\\ =x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\\ =\left(x^3+x^2+x+1\right)\left(x+1\right)\\ =\left[\left(x^3+x^2\right)+\left(x+1\right)\right]\left(x+1\right)\\ =\left[x^2\left(x+1\right)+\left(x+1\right)\right]\left(x+1\right)\\ =\left(x^2+1\right)\left(x+1\right)^2\)
\(x^4+8x=x\left(x^3+8\right)=x\left(x+2\right)\left(x^2-2x+4\right)\)
Ta có : \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
Ta có: \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
\(=\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)\)
\(=\left(2x-5\right)\left(2x+5-2x-7\right)\)
\(=-2\left(2x-5\right)\)
\(2x^2-x^2-3x-1\)
\(=x^2-3x-1\)
\(=\frac{1}{4}\left(4x^2-12x-4\right)\)
\(=\frac{-1}{4}\left[13-\left(4x-12x+9\right)\right]\)
\(=-\frac{1}{4}\left[13-\left(2x-3\right)^2\right]\)
\(=-\frac{1}{4}\left(\sqrt{13}-2x+3\right)\left(\sqrt{13}+2x-3\right)\)
TA Có : x4 + ( x -1 ) ( x2 - 2x + 1 ) + ( x - 1 )
= x4 + ( x - 1) ( x - 1 ) + ( x - 1 )
= x4 + (x -1 ) ( x - 1 + 1 )
= x4 + x ( x-1 )
= x4+ x2 - x
=x ( x3 + x -1)
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