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bài 2 :
0,25x3+x2+x=0
<=>0,25x3+0,5x2+0,5x2+x=0
<=>0,25x2(x+2)+0,5x(x+2)=0
<=>(x+2)(0,25x2+0,5x)=0
<=>(x+2)x(0,25x+0,5)=0
<=>x+2=0 hoặc x=0 hoặc 0,25x+0,5=0
=>x=-2 hoặc x=0 hoặc x=-2
vậy x=0 hoặc x=-2
a) \(xy+y^2-x-y=y\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(y-1\right)\)
b) \(25-x^2+4xy-4y^2=25-\left(x-2y\right)^2=\left(5-x+2y\right)\left(5+x-2y\right)\)
c) \(x^2-4x+3=x^2-x-3x+3=x\left(x-1\right)-3\left(x-1\right)=\left(x-1\right)\left(x-3\right)\)
d) \(y^2\left(x-1\right)-7y^3+7xy^3\)
\(=y^2\left(x-1-7y+7xy\right)\)
\(=y^2\left[\left(x-1\right)-7y\left(1-x\right)\right]=y^2\left(x-1\right)\left(1+7y\right)\)
a)
\(xy+y^2-x-y\\ =\left(xy-x\right)+\left(y^2-y\right)\\ =x\left(y-1\right)+y\left(y-1\right)\\ =\left(y-1\right)\left(x+y\right)\)
\(5x-5y+ax-ay=5\left(x-y\right)+a\left(x-y\right)=\left(5+a\right)\left(x-y\right)\)
\(a^3-a^2x-ay+xy=a^2\left(a-x\right)-y\left(a-x\right)=\left(a^2-y\right)\left(a-x\right)\)
\(10x^2+10xy+5x+5y=10x\left(x+y\right)+5\left(x+y\right)=5\left(2x+1\right)\left(x+y\right)\) \(5ay-3bx+ax-15by=a\left(5y+x\right)-3b\left(5y+x\right)=\left(a-3b\right)\left(5y+x\right)\) \(x^3+x^2-x-1=x^2\left(x+1\right)-\left(x+1\right)=\left(x^2-1\right)\left(x+1\right)=\left(x+1\right)^2\left(x-1\right)\) \(2bx-3ay-6by+ax=x\left(2b+a\right)-3y\left(2b+a\right)=\left(x-3y\right)\left(2b+a\right)\)
\(x+2a\left(x-y\right)-y=\left(x-y\right)+2a\left(x-y\right)=\left(1+2a\right)\left(x-y\right)\)
\(10x^2+10xy+5x+5y\)
\(=\left(10x^2+10xy\right)+\left(5x+5y\right)\)
\(=10x\left(x+y\right)+5\left(x+y\right)\)
\(=\left(10x+5\right)\left(x+y\right)\)
\(=5\left(5x+1\right)\left(x+y\right)\)
\(3,x\left(x-1\right)-y\left(1-x\right)=\left(x+y\right)\left(x-1\right)\\ 4,x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\\ 5,x^2-2xy+y^2-xz+yz=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y-z\right)\left(x-y\right)\\ 6,x^2-y^2-x+y=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\\ 9,x^3+x^2-xy+xy+y^2+y^3\\ =x^2\left(x+1\right)+y^2\left(x+1\right)=\left(x^2+y^2\right)\left(x+1\right)\\ 10,x^2-6\left(x+3\right)-9\\ =x^2-6x-18-9\\ =x^2-6x-27=\left(x-9\right)\left(x+3\right)\)
10: \(x^2-6\left(x+3\right)-9\)
\(=x^2-6x-18-9\)
\(=x^2-6x-27\)
\(=\left(x-9\right)\left(x+3\right)\)
a) \(x^3-2x^2+2x-1^3\)
\(=x\left(x^2-2x+1\right)+x-1\)
\(=x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x+1\right)\left(x-1\right)\)
b) \(x^2y+xy+x+1\)
\(=xy\left(x+1\right)+\left(x+1\right)\)
\(=\left(xy+1\right)\left(x+1\right)\)
c) \(ax+by+ay+bx\)
\(=a\left(x+y\right)+b\left(x+y\right)\)
\(=\left(a+b\right)\left(x+y\right)\)
d) \(x^2-\left(a+b\right)x+ab\)
\(=x^2-ax-bx+ab\)
\(=\left(x^2-ax\right)-\left(bx-ab\right)\)
\(=x\left(x-a\right)-b\left(x-a\right)\)
\(=\left(x-b\right)\left(x-a\right)\)
e) Ko biết làm
f) \(ax^2+ay-bx^2-by\)
\(=\left(ax^2+ay\right)-\left(bx^2+by\right)\)
\(=a\left(x^2+y\right)-b\left(x^2+y\right)\)
\(=\left(a-b\right)\left(x^2+y\right)\)
x^2 - xy + x - y = x(x - y) + (x - y) = (x - y)(x + 1)
1.Phân tích thành nhân tử ( phương pháp nhóm nhiều hạng tử )
a. x^3 + 2x^2 - xy - 2y
\(=x^2\left(x+2\right)-y\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-y\right)\)
b. xy - 5x + 3y^2 - 15y
\(=xy+3y^2-5x-15y\)
\(=y\left(x+3y\right)-5\left(x+3y\right)\)
\(=\left(x+3y\right)\left(y-5\right)\)
c.2xy + 6x + y^2 + 3y
\(=2xy+y^2+6x+3y\)
\(=y\left(2x+y\right)+3\left(2x+y\right)\)
\(=\left(2x+y\right)\left(y+3\right)\)
a) \(x^3+2x^2-xy-2y\)
\(=\left(x^3-xy\right)+\left(2x^2-2y\right)\)
\(=x\left(x^2-y\right)+2\left(x^2-y\right)\)
\(=\left(x+2\right)\left(x^2-y\right)\)
\(=\left(x+2\right)\left(x+\sqrt{y}\right)\left(x-\sqrt{y}\right)\)
xin loi bn nha
mk moi chi hok lop 6
mk nghi bn nen tu lam thi hon,ko nen dua vao olm.vn qua
chuc bn hok tot nha
2/\(5ay-3bx+ax-15by.\)
\(=\left(5ay-15by\right)-\left(3bx-ax\right)\)
\(=5y\left(3ab\right)-x\left(3ab\right)\)
\(=\left(3ab\right)\left(5y-x\right)\)
3/\(x^2+xy+x+1\)
\(=xy+x^2+x+1\)
4/\(x^2y+xy^2-x-y\)
\(=xy\left(x+y\right)-x-y\)
=.= mik ko chắc