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a) \(\left(x+y\right)^3-x^3-y^3\)
\(=\left(x+y\right)^3-\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-x^2+xy-y^2\right]\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-x^2+xy-y^2\right)\)
\(=3xy\left(x+y\right)\)
b) \(x^2+y^2+2xy+yz+xz\)
\(=\left(x^2+2xy+y^2\right)+\left(yz+xz\right)\)
\(=\left(x+y\right)^2+z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y+z\right)\)
c) \(x^2-10xy-1+25y^2\)
\(=\left(x^2-10xy+25y^2\right)-1\)
\(=\left(x-5y\right)^2-1\)
\(=\left(x-5y-1\right)\left(x-5y+1\right)\)
d) \(ax^2-ax+bx^2-bx+a+b\)
\(=(ax^2+bx^2)-(ax+bx)+(a+b)\)
\(=x^2(a+b)-x(a+b)+(a+b)\)
\(=(a+b)(x^2-x+1)\)
e)\(x^2-2y+3xz+x-2y+3z\)
\(=(x^2+x)-(2xy+2y)+(3xz+3z)\)
\(=x(x+1)-2y(x-1)+3z(x+1)\)
\(=(x+1)(x-2y+3z)\)
f) \(xyz-xy-yz-xz+x+y+z-1\)
\(=(xyz-xy)-(yz-y)-(xz-x)+(z-1)\)
\(=xy(z-1)-y(z-1)-x(z-1)+(z-1)\)
\(=(z-1)(xy-y-x+1)\)
\(=(z-1)[y(x-1)-(x-1)]\)
\(=(z-1)(x-1)(y-1)\)
_Học tốt_
c) \(x^2+y^2+xz+yz+2xy\)
\(=\left(x+y\right)^2+z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y+z\right)\)
b) \(x^3+3x^2-3x-1\)
\(=\left(x^3-1\right)+3x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+4x+1\right)\)
a) \(g\left(x,y\right)=x^2-10xy+9y^2=x^2-xy-9xy+9y^2\)
\(=x\left(x-y\right)-9y\left(x-y\right)=\left(x-y\right)\left(x-9y\right)\).
b )\(f\left(x,y\right)=x^6+x^4+x^2y^2+y^4-y^6\)
\(=x^6-y^6+x^4+x^2y^2+y^4\)
\(=\left(x^3\right)^2-\left(y^3\right)^2+\left(x^4+2x^2y^2+y^4\right)-x^2y^2\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)+\left(x^2+y^2\right)^2-\left(xy\right)^2\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)+\left(x^2+y^2-xy\right)\left(x^2+y^2+xy\right)\)
\(=\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\left[\left(x-y\right)\left(x+y\right)+1\right]\)
\(=\left(x^2+xy+y^2\right)\left(x^2-2y+y^2\right)\left(x^2-y^2+1\right)\)
Vậy \(f\left(x,y\right)=\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\left(x^2-y^2+1\right)\)
a)xz-yz -x2 +2xy-y2=(xz-yz)-(x2-2xy+y2)=z(x-y)-(x-y)2=(x-y)(z-x+y)
b) x2+8x+15= (x2+3x)+(5x+15)=x(x+3)+5(x+3)=(x+3)(x+5)
c) x2-x-12=(x2-4x)+(3x-12)=x(x-4)+3(x-4)=(x-4)(x+3)
a) xz - yz - x2 + 2xy - y2
= (xz - yz) - (x2 - 2xy + y2)
= z (x - y) - (x - y)2
= z (x - y) - (x - y) (x - y)
= [z - (x - y)] (x - y)
= (z - x + y) (x - y)
b) x2 + 8x + 15
= x2 + 3x + 5x + 15
= (x2 + 3x) + (5x + 15)
= x (x + 3) + 5 (x + 3)
= (x + 5) (x + 3)
c) x2 - x - 12
= x2 - 4x + 3x - 12
= (x2 - 4x) + (3x - 12)
= x (x - 4) + 3 (x - 4)
= (x + 3) (x - 4)
#Học tốt!!!
~NTTH~
\(x^2-2xy+y^2-xz+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
Bài làm
a) x2 - 2xy + y2 - zx + yz
= ( x2 - 2xy + y2 ) - ( zx - yz )
= ( x - y )2 - z( x - y )
= ( x - y )( x - y - z )
b) x3 - x2 - 5x + 125
= ( x3 + 125 ) - ( x2 + 5x )
= ( x + 5 )( x2 -.5x + 25 ) - x( x + 5 )
= ( x + 5 )( x2 - 5x + 25 - x )
= ( x + 5 )( x2 - 6x + 25 )
# Học tốt #
câu a nhầm đề à bạn,mk nghĩ -xz chứ ko phải -xy.
Câu 2 nha
\(a,x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
\(c,x^2-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
Phân tích đa thức thành nhân tử
a) 125x3 + y6
= (5x)3 + (y2)3
= (5x + y2)(25x2 - 5xy2 + y4)
b) x2 - 2xy + y2 - xz + yz
= (x - y)2 - z(x - y)
= (x - y)(x - y - z)
c) x4 - x3 - x2 + 1
= (x4 - x3) - (x2 - 1)
= x3(x - 1) - (x - 1)(x + 1)
= (x - 1)(x3 - x + 1)
uk