phân tích đa thức thành nhân tử : 2x ^ 2 -5 xy -3y ^ 2
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
^2 + 4xy - 16 + 4y^2
= x^2 + 4xy + 4y^2 - 4^2
= (x + 2y)^2 - 4^2
= (x + 2y - 4)(x + 2y + 4)
2x^2-5xy-3y^2
= 2^x + xy - 6xy - 3y^2
= x(2x + y) - 3y(2x + y)
= (2x + y)(x - 3y)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$A=x^2-8xy+15y^2=x^2-3xy-(5xy-15y^2)$
$=x(x-3y)-5y(x-3y)=(x-3y)(x-5y)$
$B=2x^2-5xy+2y^2=(2x^2-4xy)-(xy-2y^2)$
$=2x(x-2y)-y(x-2y)=(2x-y)(x-2y)$
$C=2x^2-3y^2-xy=(2x^2+2xy)-(3y^2+3xy)$
$=2x(x+y)-3y(y+x)=(x+y)(2x-3y)$
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(2\left(x-1\right)^3-\left(x-1\right)=\left(x-1\right)\left(2\left(x-1\right)^2-1\right)\)
2) \(y\left(x-2y\right)^2+xy^2\left(2y-x\right)=\left(2y-x\right)\left(2\left(2y-x\right)+1\right)=\left(2y-x\right)\left(4y-2x+1\right)\)
3) \(xy\left(x+y\right)-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\) (xem lại đề sửa -2x thành -x mới đúng)
4) \(xy\left(x-3y\right)-2x+6y=xy\left(x-3y\right)-2\left(x-3y\right)=\left(x-3y\right)\left(xy-2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x2 + 3x + 3y + xy
= ( x2 + xy) + ( 3x + 3y)
= x( x + y) + 3 ( x + y)
= ( x + y) ( x + 3)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left(2x+3y\right)^2-2\left(2x+3y\right)\)
\(=\left(2x+3y\right)\left(2x+3y-2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(xy^3-5x^2\)
\(=x\cdot y^3-x\cdot5x\)
\(=x\left(y^3-5x\right)\)
\(y^2+3y\)
\(=y\cdot y+y\cdot3\)
\(=y\left(y+3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3y-xy^3-2xy^2-xy\)
\(=xy\left(x^2-y^2-2y-1\right)\)
\(=xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=xy\left[x^2-\left(y+1\right)^2\right]\)
\(=xy\left(x-y-1\right)\left(x+y+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x+3y-x^2-xy\)
\(=\left(3x+3y\right)-\left(x^2+xy\right)\)
\(=3\left(x+y\right)-x\left(x+y\right)\)
\(=\left(3-x\right)\left(x+y\right)\)
\(2x^2-5xy-3y^2\)
=\(2x^2+xy-6xy-3y^2\)
\(=x\left(2x+y\right)-3y\left(2x+y\right)\)
=\(\left(x-3y\right)\left(2x+y\right)\)