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![](https://rs.olm.vn/images/avt/0.png?1311)
a,x^2-x-y^2-y
=x^2-y^2-(x+y)
=(x-y).(x+y)-(x+y)
=(x+y).(x-y-1)
b, x^2-2xy+y^2-z^2
=(x^2-2xy+y^2)-z^2
=(x-y)^2-z^2
=(x-y-z)(x-y+z)
c,5x-5y+ax-ay( đề bài ở đây phải là -ay ms tính đc)
=(5x-5y)+(ax-ay)
=5(x-y)+a(x-y)
=(x-y).(5+a)
d,a^3-a^2.x-ay+xy
=(a^3-a^2x)-(ay-xy)
=a^2(a-x)-y(a-x)
=(a-x)(a^2-y)
e,4x^2-y^2+4x+1
={(2x)^2+4x+1}-y^2
=(2x+1)^2-y^2
=(2x+1+y^2)(2x+1-y^2)
f,x^3-x+y^3-y
=(x^3+y^3)-(x+y)
=(x+y)(x^2-xy+y^2)-(x+y)
=(x+y)(x^2-xy+y^2-1)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) x2 - x - y2 - y = (x - y)(x + y) - (x + y) = (x - y - 1)(x + y)
2. x2 - 2xy + y2 - z2 = (x - y)2 - z2 = (x - y - z)(x - y + z)
3. 5x - 5y + ax - ay = 5(x - y) + a(x - y) = (a + 5)(x - y)
4. a3 - a2x - ay + xy = a2(a - x) - y(a - x) = (a2 - y)(a - x)
5. 4x2 - y2 + 4x + 1 = (2x + 1)2 - y2 = (2x + 1 - y)(2x + y + 1)
6. x3 - x + y3 - y = (x + y)(x2 - xy + y2) - (x + y) = (x + y)(x2 - xy + y2 - 1)
Trả lời:
1, x2 - x - y2 - y
= ( x2 - y2 ) - ( x + y )
= ( x - y ) ( x + y ) - ( x + y )
= ( x + y ) ( x - y - 1 )
2, x2 - 2xy + y2 - z2
= ( x2 - 2xy + y2 ) - z2
= ( x - y )2 - x2
= ( x - y - z ) ( x - y + z )
3, 5x - 5y + ax - ay
= ( 5x + ax ) - ( 5y + ay )
= x ( 5 + a ) - y ( 5 + a )
= ( 5 + a ) ( x - y )
= ( 5 + a ) ( x - y )
4, a3 - a2x - ay + xy
= ( a3 - a2x ) - ( ay - xy )
= a2 ( a - x ) - y ( a - x )
= ( a - x ) ( a2 - y )
5, 4x2 - y2 + 4x + 1
= ( 4x2 + 4x + 1 ) - y2
= ( 2x + 1 )2 - y2
= ( 2x + 1 - y ) ( 2x + 1 + y )
6, x3 - x + y3 - y
= ( x3 + y3 ) - ( x + y )
= ( x + y ) ( x2 - xy + y ) - ( x + y )
= ( x + y ) ( x2 - xy + y - 1 )
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+2xy+y^2-xz-yz\)
\(=\left(x+y\right)^2-z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
mk chỉnh lại đề
\(x^2-2xy+y^2-z^2+2zt+t^2\)
\(=\left(x-y\right)^2-\left(z-t\right)^2\)
\(=\left(x-y-z+t\right)\left(x-y+z-t\right)\)
mk chỉnh lại đề:
\(ax^2+cx^2-ay+ay^2-cy+cy^2\)
\(=x^2\left(a+c\right)-y\left(a+c\right)+y^2\left(a+c\right)\)
\(=\left(a+c\right)\left(x^2-y+y^2\right)\)
\(ax^2+ay^2-bx^2-by^2+b-a\)
\(=x^2\left(a-b\right)+y^2\left(a-b\right)-\left(a-b\right)\)
\(=\left(a-b\right)\left(x^2+y^2-1\right)\)
\(ac^2-ad-bc^2+cd+bd-c^3\)
\(=a\left(c^2-d\right)-b\left(c^2-d\right)-c\left(c^2-d\right)\)
\(=\left(c^2-d\right)\left(a-b-c\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,=\left(x-3\right)\left(x+3\right)\\ 2,=\left(x-y\right)\left(5+a\right)\\ 3,=\left(x+3\right)^2\\ 4,=\left(x-y\right)\left(10x+7y\right)\\ 5,=5\left(x-3y\right)\\ 6,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ax+bx+ay+by\)
\(=x\left(a+b\right)+y\left(a+b\right)\)
\(=\left(x+y\right)\left(a+b\right)\)
\(xy+1-x-y\)
\(=x\left(y-1\right)-\left(y-1\right)\)
\(=\left(x-1\right)\left(y-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1 \(=\left(4x^2+4x+1\right)-\left(3y\right)^2\)
\(=\left(2x+1\right)^2-\left(3y\right)^2\)
\(=\left(2x+1-3y\right)\left(2x+1+3y\right)\)
2,\(=\left(x^2+2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)
\(=\left(x+y\right)^2-\left(z-t\right)^2\)
\(=\left(x+y+z-t\right)\left(x+y-z+t\right)\)
3,\(=9x\left(x-y\right)-7\left(x-y\right)\)
\(=\left(x-y\right)\left(9x-7\right)\)
4\(=3\left(x-y\right)+a\left(x-y\right)\)
\(=\left(x-y\right)\left(3+a\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1 ) \(x^2-x-y^2-y=\left(x^2-y^2\right)+\left(-x-y\right)=\left(x+y\right)\left(x-y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)
2 ) \(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y+z\right)\left(x-y-z\right)\)
3 ) \(5x-5y+ax-ay=5.\left(x-y\right)+a\left(x-y\right)=\left(x-y\right)\left(5+a\right)\)
4 ) \(a^3-a^2x-ay+xy=a^2.\left(a-x\right)-y.\left(a-x\right)=\left(a-x\right)\left(a^2-y\right)\)
5 ) \(xy.\left(x+y\right)+yz.\left(y+z\right)+xz.\left(x+z\right)+2xyz\)
\(=xy.\left(x+y\right)+y^2z+yz^2+x^2z+xz^2+xyz+xyz\)
\(=xy.\left(x+y\right)+\left(y^2z+xyz\right)+\left(yz^2+xz^2\right)+\left(x^2z+xyz\right)\)
\(=xy.\left(x+y\right)+yz.\left(x+y\right)+z^2.\left(x+y\right)+xz.\left(x+y\right)\)
\(=\left(x+y\right)\left(xy+yz+z^2+xz\right)=\left(x+y\right)\left[\left(xy+xz\right)+\left(yz+z^2\right)\right]\)
\(=\left(x+y\right)\left[x.\left(y+z\right)+z.\left(y+z\right)\right]=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)