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x2+4xy+2x+3y2+6y
=(x2+xy+2x)+(3xy+3y2+6y)
=x(x+y+2)+3y(x+y+2)
=(x+y+2)(x+3y)
Áp dụng \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)
\(=x^3+y^3+3xy\left(x+y\right)+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(2x^4+x^3-22x^2+15x-36\)
\(=\left(2x^4-6x^3\right)+\left(7x^3-21x^2\right)+\left(-x^2+3x\right)+\left(12x-36\right)\)
\(=\left(x-3\right)\left(2x^3+7x^2-x+12\right)\)
\(=\left(x-3\right)\left(\left(2x^3+8x^2\right)+\left(-x^2-4x\right)+\left(3x+12\right)\right)\)
\(=\left(x-3\right)\left(x+4\right)\left(2x^2-x+3\right)\)
x^8 + x + 1 = (x^8 + x^7 + x^6) - ( x^7 + x^6 + x^5) + (x^5 + x^4 + x^3) -(x^4 + x^3 + x^2) + (x^2+x+1)
= (x^2+x+1)(x^6 - x^5 + x^3 - x^2 +1)
Ta có \(x^8+x+1=x^8-x^2+x^2+x+1\)
\(=x^2\left(x^6-1\right)+x^2+x+1\)
\(=x^2\left(x^3-1\right)\left(x^3+1\right)+x^2+x+1\)
\(=\left(x^3-1\right)\left(x^5+x^2\right)+x^2+x+1\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x^5+x^2\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\text{[}\left(x-1\right)\left(x^5+x^2\right)+1\text{]}\text{ }\)
\(=\left(x^2+x+1\right)\left(x^6+x^3-x^5-x^2+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
Vậy \(x^8+x+1=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)
x3-5x2+2x+8
=x3-6x2+8x+x2-6x+8
=x(x2-6x+8)+(x2-6x+8)
=(x2-6x+8)(x+1)
=[x2-2x-4x+8](x+1)
=[x(x-2)-4(x-2)](x+1)
=(x-4)(x-2)(x+1)