Phân tích đa thức sau thành nhân tử:
2x3-3x2+3x-1
Giúp mình nha.Ai nhanh mình k cho. ^^
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a) (x - y)(x + y + 3). b) (x + y - 2xy)(2 + y + 2xy).
c) x 2 (x + l)( x 3 - x 2 + 2). d) (x – 1 - y)[ ( x - 1 ) 2 + ( x - 1 ) y + y 2 ].
1/(x+2)2 -(3x-1)2=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x2+12x
2/(x4+x2)(-2x3-2x)=x2(x2+1)-2x(x2+1)=(x2+1)(x2-2x)
\(=x\left(2x^2+3x-2\right)=x\left(2x^2+4x-x-2\right)=x\left[2x\left(x+2\right)-\left(x+2\right)\right]=x\left(2x-1\right)\left(x+2\right)\)
2x3 + 3x2 - 2x
= x ( 2x2 + 3x - 2 )
= x ( 2\(x^2\) + 4\(x-x-2\) )
= x [ ( 2\(x^2\) + 4x ) - ( x + 2 )]
= x [ 2x ( x + 2 ) - ( x + 2 )]
= x ( 2x - 1 ) ( x + 2 )
1a) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
b) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)
\(\frac{\left(a-b\right)^2}{4}\)- 1 = (\(\frac{a-b}{2}\)- 1)(\(\frac{a-b}{2}\)+ 1)
d) x3 + 3x2 – 3x – 1
= (x3 - 1) + (3x2 - 3x)
= (x - 1)(x2 + x + z) + 3x(x - 1)
= (x - 1)(x2 + 4x + 1)
a) x3 + 3x2 – 3x – 9
= (x3 + 3x2) - (3x + 9)
= x2(x + 3) - 3(x + 3)
= (x + 3)(x2 - 3)
= (x + 3)(x + √3)(x - √3)
\(2x^3-3x^2+3x-1=x^3+x^3-3x^2+3x-1\)
=\(x^3+\left(x^3-3x^2+3x-1\right)\)=\(x^3+\left(x-1\right)^3\)
=\(\left(x+x-1\right)\left(x^2-x\left(x-1\right)+\left(x-1\right)^2\right)\)
=\(\left(2x-1\right)\left(x^2-x^2+x+x^2-2x+1\right)\)
=\(\left(2x-1\right)\left(x^2-x+1\right)\)