Bài 1: Phân tích đa thức thành nhân tử
a) \(y^2.\left(z^2+y\right)-x^2z-zy\)
b) \(x^4-x^3-x^2-1\)
c) \(ax^2+a^2y-7x-7y\)
d) \(x^2y+xy^2+-x-y\)
e) \(ax^2+ay-bx^2-by\)
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a)x2-xy-x+y
=(x2-x)-(xy-y)
=x(x-1)-y(x-1)
=(x-1)(x-y)
b) xy+4-x2+2y
=(4-x2)+(xy+2y)
=(2-x)(x+2)+y(x+2)
=(x+2)(2-x+y)
c) xy+y-2(x+1)
=y(x+1)-2(x+1)
=(x+1)(y-2)
d) 5(x-y)+ax-ay
=5(x-y)+a(x-y)
=(x-y)(5+a)
#H
Trả lời:
a, x2 - xy - x + y
= ( x2 - xy ) - ( x - y )
= x ( x - y ) - ( x - y )
= ( x - y ) ( x - 1 )
b, xy + 4 - x2 + 2y
= ( xy + 2y ) - ( x2 - 4 )
= y ( x + 2 ) - ( x - 2 ) ( x + 2 )
= ( x + 2 ) ( y - x + 2 )
c, xy + y - 2 ( x + 1 )
= y ( x + 1 ) - 2 ( x + 1 )
= ( x + 1 ) ( y - 2 )
d, 5 ( x - y ) + ax - ay
= 5 ( x - y ) + a ( x - y )
= ( 5 + a ) ( x - y )
\(a,ax+by+ay+bx=\left(ax+ay\right)+\left(by+bx\right)=a\left(x+y\right)+b\left(x+y\right)=\left(a+b\right)\left(x+y\right)\)
\(b,x^2y+xy+x+1=xy\left(x+1\right)+\left(x+1\right)=\left(xy+1\right)\left(x+1\right)\)
\(c,x^2-ax-bx+ab=x\left(x-a\right)-b\left(x-a\right)=\left(x-b\right)\left(x-2\right)\)
\(d,x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(xy-1\right)\left(x+y\right)\)
\(e,a\left(x^2+y\right)-b\left(x^2+y\right)=\left(a-b\right)\left(x^2+y\right)\)
\(f,x\left(a-2\right)-a\left(a-2\right)=\left(x-a\right)\left(a-2\right)\)
bn post nhiều nên mình ghi đáp án thôi nhé phần nào sai đề mình cho qua
b)\(\left(x+1\right)\left(xy+1\right)\)
c)\(\left(a+b\right)\left(x+y\right)\)
d)\(\left(x-a\right)\left(x-b\right)\)
e)\(\left(x+y\right)\left(xy-1\right)\)
f)\(\left(a-b\right)\left(x^2+y\right)\)
a) bạn ktra lại đề
b) \(x^2y+xy+x+1=xy\left(x+1\right)+\left(x+1\right)=\left(xy+1\right)\left(x+1\right)\)
c) \(ax+by+ay+bx=a\left(x+y\right)+b\left(x+y\right)=\left(a+b\right)\left(x+y\right)\)
d) \(x^2-\left(a+b\right)x+ab=x^2-ax-bx+ab=x\left(x-a\right)-b\left(x-a\right)=\left(x-a\right)\left(x-b\right)\)
e) \(x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(xy-1\right)\left(x+y\right)\)
f) \(ax ^2+ay-bx^2-by=x^2\left(a-b\right)+y\left(a-b\right)=\left(a-b\right)\left(x^2+y\right)\)
\(x^2y+xy+x+1\)
\(=xy\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(xy+1\right)\)
hk tốt
^^