Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) xz-yz+5y-5x=\(z\left(x-y\right)+5\left(y-x\right)\)=\(z\left(x-y\right)-5\left(x-y\right)\)=\(\left(z-5\right)\left(x-y\right)\)
b) \(3x^2-6x+3-3y^2\)=\(3\left(x^2-2x+1-y^2\right)\)=\(3\left(\left(x-1\right)^2-y^2\right)\)=\(3\left(x-1-y\right)\left(x-1+y\right)\)
\(a,=\left(xy-1-x-y\right)\left(xy-1+x+y\right)\\ b,Sửa:a^3+2a^2+2a+1\\ =a^3+a^2+a^2+a+a+1=\left(a+1\right)\left(a^2+a+1\right)\\ c,=1-4a^2-a\left(a^2-4\right)=1-4a^2-a^3+4a\\ =\left(1-a\right)\left(1+a+a^2\right)+4a\left(1-a\right)\\ =\left(1-a\right)\left(1+5a+a^2\right)\\ d,=\left(a^2-a^2b^2\right)+\left(b^2-b\right)+\left(ab-a\right)\\ =a^2\left(1-b\right)\left(1+b\right)+b\left(b-1\right)+a\left(b-1\right)\\ =\left(b-1\right)\left(-a^2-ab+b+a\right)\\ =\left(b-1\right)\left(b-1\right)\left(a+b\right)\left(1-a\right)\)
\(e,=x^2y+xy^2-yz\left(y+z\right)+x^2z-xz^2\\ =\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\\ =x^2\left(y+z\right)+x\left(y-z\right)\left(y+z\right)-yz\left(y+z\right)\\ =\left(y+z\right)\left(x^2+xy-xz-yz\right)\\ =\left(y+z\right)\left(x+y\right)\left(x-z\right)\)
\(f,=xyz-xy-yz-xz+x+y+z-1\\ =xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(x-1\right)\\ =\left(z-1\right)\left(xy-y-x+1\right)=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)
1) \(4x^2-7x-2=4x^2-8x+x-2=\left(4x^2-8x\right)+\left(x-2\right)\)
\(=4x\left(x-2\right)+\left(x-2\right)=\left(x-2\right)\left(4x+1\right)\)
2) \(4x^2+5x-6=4x^2+8x-3x-6=\left(4x^2+8x\right)-\left(3x+6\right)\)
\(=4x\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(4x-3\right)\)
3) \(5x^2-18x-8=5x^2-20x+2x-8=\left(5x^2-20x\right)+\left(2x-8\right)\)
\(=5x\left(x-4\right)+2\left(x-4\right)=\left(x-4\right)\left(5x+2\right)\)
4) \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)
\(=xy\left(x+y\right)-y^2z-yz^2+x^2z-xz^2\)
\(=xy\left(x+y\right)+\left(x^2z-y^2z\right)-\left(yz^2+xz^2\right)\)
\(=xy\left(x+y\right)+z\left(x^2-y^2\right)-z^2.\left(x+y\right)\)
\(=xy\left(x+y\right)+z\left(x-y\right)\left(x+y\right)-z^2\left(x+y\right)\)
\(=xy\left(x+y\right)+\left(zx-zy\right)\left(x+y\right)-z^2\left(x+y\right)\)
\(=\left(x+y\right)\left(xy+xz-yz-z^2\right)=\left(x+y\right).\left[x\left(y+z\right)-z\left(y+z\right)\right]\)
\(=\left(x+y\right)\left(y+z\right)\left(x-z\right)\)
1) 4x2 - 7x - 2 = 4x2 - 8x + x - 2 = 4x( x - 2 ) + ( x - 2 ) = ( x - 2 )( 4x + 1 )
2) 4x2 + 5x - 6 = 4x2 - 8x + 3x - 6 = 4x( x - 2 ) + 3( x - 2 ) = ( x - 2 )( 4x + 3 )
3) 5x2 - 18x - 8 = 5x2 - 20x + 2x - 8 = 5x( x - 4 ) + 2( x - 4 ) = ( x - 4 )( 5x + 2 )
4) xy( x + y ) - yz( y + z ) + xz( x - z )
= x2y + xy2 - y2z - yz2 + xz( x - z )
= ( x2y - yz2 ) + ( xy2 - y2z ) + xz( x - z )
= y( x2 - z2 ) + y2( x - z ) + xz( x - z )
= y( x - z )( x + z ) + y2( x - z ) + xz( x - z )
= ( x - z )[ y( x + z ) + y2 + xz ]
= ( x - z )( xy + yz + y2 + xz )
= ( x - z )[ ( xy + y2 ) + ( xz + yz ) ]
= ( x - z )[ y( x + y ) + z( x + y ) ]
= ( x - z )( x + y )( y + z )
5) xy( x + y ) + yz + xz( x + z ) + 2xyz ( đề có thiếu không vậy .-. )
Trả lời:
a, \(27a^2b^2-18ab+3=3\left(9a^2b^2-6ab+1\right)=3\left(3ab-1\right)^2\)
b, \(x^2+2xy+y^2-xz-yz\)
\(=\left(x^2+2xy+y^2\right)-z\left(x+y\right)\)
\(=\left(x+y\right)^2-z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
c, \(a^4+a^3-a^2-a\)
\(=\left(a^4+a^3\right)-\left(a^2+a\right)\)
\(=a^3\left(a+1\right)-a\left(a+1\right)\)
\(=a\left(a+1\right)\left(a^2-1\right)\)
\(=a\left(a+1\right)\left(a-1\right)\left(a+1\right)\)
\(=a\left(a+1\right)^2\left(a-1\right)\)
d, \(a^3-b^3+2b-2a\)
\(=\left(a^3-b^3\right)-\left(2a-2b\right)\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)-2\left(a-b\right)\)
\(=\left(a-b\right)\left(a^2+ab+b^2-2\right)\)
1) \(x^2-2xy+y^2-xz+yz\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)-\left(xz-yz\right)\)
\(\Leftrightarrow\left(x-y\right)^2-z\left(x-y\right)\)
\(\Leftrightarrow\left(x-y\right)\left(x-y-z\right)\)
2)\(x^2-y^2-x+y\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-\left(x-y\right)\)
\(\Leftrightarrow\left(x-y\right)\left(x+y+1\right)\)
\(a,x^2-2xy+y^2-xz+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
\(b,x^2-y^2-x+y\)
\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-1\right)\)
a.\(xz+yz-5\left(x+y\right)\)
\(=z\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(z-5\right)\)
b.\(3x^2-3xy-5x+5y\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
c.\(x^2+6x-y^2-3z^2\)???Sai đề bài ...?
d.\(3x^2+6xy+3y^2-3z^2\)
\(=3\left(x^2+2xy+y^2-z^2\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]\)'
\(=3\left(x+y-z\right)\left(x+y+z\right)\)
Trả lời:
a, xz + yz - 5 ( x + y )
= ( xz + yz ) - 5 ( x + y )
= z ( x + y ) - 5 ( x + y )
= ( x + y ) ( z - 5 )
b, 3x2 - 3xy - 5x + 5y
= ( 3x2 - 3xy ) - ( 5x - 5y )
= 3x ( x - y ) - 5 ( x - y )
= ( x - y ) ( 3x - 5 )
c, x2 + 6x - y2 - 3z2
= - ( 3x2 - x2 + y2 - 6x )
d, 3x2 + 6xy + 3y2 - 3z2
= 3 ( x2 + 2xy + y2 - x2 )
= 3 [ ( x2 + 2xy + y2 ) - z2 ]
= 3 [ ( x + y )2 - z2 ]
= 3 ( x + y - z ) ( x + y + z )
\(a^2-8a+15\)
\(=\left(a^2-2.4a+4^2\right)-1^2\)
\(=\left(a-4\right)^2-1^2\)
\(=\left(a-4-1\right)\left(a-4+1\right)\)
\(=\left(a-3\right).\left(a-5\right)\)
\(a^2-8a+15\)
\(=a^2-2.a.4+16-1\)
\(=\left(a-4\right)^2-1\)
\(=\left(a-4-1\right)\left(a-4+1\right)\)
\(=\left(a-5\right)\left(a-3\right)\)
\(3x^2-10x-8\)
\(=3x^2-12x+2x-8\)
\(=3x\left(x-4\right)+2\left(x-4\right)\)
\(=\left(3x+2\right)\left(x-4\right)\)
\(-6x^3+18x^2+60x\)
\(=\)\(-6x^3+30x^2-12x^2+60x\)
\(=-6x^2\left(x-5\right)-12x\left(x-5\right)\)
\(=\)\(\left(-6x^2-12x\right)\left(x-5\right)\)
\(=-6x\left(x+2\right)\left(x-5\right)\)