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\(x^5+x^4+2\)
\(=x^5+x^4+x^2-x^2+1+1\)
\(=\left(x^5-x^2\right)+\left(x^4+x^2+1\right)\)
\(=\left(x^5-x^2\right)+\left(x^4+2x^2-x^2+1\right)+1\)
\(=x^2\left(x^3-1\right)+\left(x^4+2x^2-x^2+1\right)+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(\left(x^2+1\right)^2-x^2\right)+1\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+1+x\right)\cdot\left(x^2+1-x\right)+1\)
\(=\left(x^3-x^2\right)\left(x^2+x+1\right)+\left(x^2+1+x\right)\cdot\left(x^2+1-x\right)+1\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+x^2+1-x\right)+1\)
\(=\left(x^2+x+1\right)\left(x^3+1-x\right)+1\)
x2 + 1 - y2 - 2x
= x2 - 2x + 1 - y2
=[x2 - 2x + 1] - y2
=[x-1]2 - y2
=[x-1-y][x-1+y]
a) \(x^2+1-y^2-2x=\left(x^2-2x+1\right)-y^2=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)
b) \(64x^4+y^4=\left(8x^2\right)^2+\left(y^2\right)^2=\left(8x^2\right)^2+16x^2y^2+\left(y^2\right)^2-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2=\left(8x^2+y^2-4xy\right)\left(8x^2+y^2+4xy\right)\)
Nghịch xíu :v
a, \(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)-2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-2x+2\right)\)
b, \(x^2+4x+3\)
\(=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
Chúc bạn học tốt!!!
a )
b)
c) x^5 - x^4 - 1
= x^5 - x^3 - x² - x^4 + x² + x + x^3 - x - 1
= x²( x^3 - x - 1 ) - x( x^3 - x - 1 ) + ( x^3 - x - 1 )
= ( x² - x + 1)( x^3 - x - 1 )
d)
=x3(x+2)-13x2+12x-26x+24
=x3(x+2)-x(13x-12)-2(13x-12)
=x3(x+2)-(13x-12)(x+2)
=(x+2)(x3-x-12x+12)
(x+2)[(x2-1)-12(x-1)]
=(x+2)[x(x-1)(x+1)-12(x-1)]
=(x+2)(x-1)[x(x+1)-12]
=(x+2)(x-1)(x2+x-12)
=(x+2)(x-1)(x2-3x+4x-12)
=(x+2)(x-1)[x(x-3)+4(x+3)]
=(x+2)(x-1)(x-3)(x+4)
trong bài làm của mk có hàng k có dấu "=" chỗ đó có dâu"=" nha!
a) \(3x^2-9x+30=3\left(x^2-3x+10\right)\)
b) \(3x^2-5x-2=3x^2-6x+x-2\)
\(=3x\left(x-2\right)+\left(x-2\right)=\left(3x+1\right)\left(x-2\right)\)
c) \(x^4+4y^4\)
\(=x^4+4y^4+2x^2y^2+2x^2y^2-4x^2y^2+4xy^3-4xy^3+2x^3y-2x^3y\)
\(=\left(4y^4-4xy^3+2x^2y^2\right)+\left(4xy^3-4x^2y^2+2x^3y\right)\)
\(+\left(2x^2y^2-2x^3y+x^4\right)\)
\(=2y^2\left(2y^2-2xy+x^2\right)+2xy\left(2y^2-2xy+x^2\right)\)
\(+x^2\left(2y^2-2xy+x^2\right)\)
\(=\left(2y^2+2xy+x^2\right)\left(2y^2-2xy+x^2\right)\)
d) \(x^5+x+1\)
\(=x^5+x+1+x^4-x^4+x^3-x^3+x^2-x^2\)
\(=\left(x^5-x^4+x^2\right)+\left(x^4-x^3+x\right)+\left(x^3-x^2+1\right)\)
\(=x^2\left(x^3-x^2+1\right)+x\left(x^3-x^2+1\right)+\left(x^3-x^2+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
\(x^3+2x^2+2x+1=\left(x^3+1\right)+\left(2x^2+2x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1+2x\right)=\left(x+1\right)\left(x^2+x+1\right)\)
\(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27\)
\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
\(x^4+2x^3+2x^2+2x+1=x^4+x^2+2x^3+x^2+2x+1\)
\(=x^2\left(x^2+1\right)+2x\left(x^2+1\right)+\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^2+2x+1\right)\)
\(=\left(x^2+1\right)\left(x+1\right)^2\)
\(x^4-2x^3+2x-1=\left(x^4-1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+1-2x\right)=\left(x^2-1\right)\left(x-1\right)^2\)
\(x^3+2x^2+2x+1=\left(x^3+x^2\right)+\left(x^2+x\right)+\left(x+1\right)\)
\(=x^2.\left(x+1\right)+x.\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right).\left(x^2+x+1\right)\)
\(x^3-4x^2+12x-27\)
\(=\left(x^3-x^2\right)-\left(3x^2-3x\right)+\left(9x-27\right)\)
\(=x^2.\left(x-1\right)-3x.\left(x-1\right)+9.\left(x-3\right)\)
\(=\left(x-1\right).\left(x^2-3x\right)+9.\left(x-3\right)\)
\(=x.\left(x-1\right).\left(x-3\right)+9.\left(x-3\right)\)
\(=\left(x-3\right)\left[x.\left(x-1\right)+9\right]\)
<=>x4-x+x2 +x+1= x (x-1) (x2+x+1) + (x2+x+1) = (x2+x+1)(x2-x+1)
chắc có lẽ đúng đó
\(x^4-x^2+1\)
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