phân tích đa thức thành nhân tử 2x^2 - 5xy - 3y^2
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
a/ \(\left(x+3y\right)\left(2x-y\right)\)
b/ \(\left(x^2+2\right)\left(x^2-2\right)\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
b/ \(x^8-16=\left(x^4+4\right)\left(x^4-4\right)\)
\(=\left[\left(x^4+4x^2+4\right)-4x^2\right]\left(x^2-2\right)\left(x^2+2\right)\)
\(=\left[\left(x^2+2\right)^2-4x^2\right]\left(x^2-2\right)\left(x^2+2\right)\)
\(=\left(x^2+2+2x\right)\left(x^2+2-2x\right)\left(x^2-2\right)\left(x^2+2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2x^2-5xy-3y^2\)
\(A=2x^2-6xy+xy-3y^2\)
\(A=2x\left(x-3y\right)+y\left(x-3y\right)\)
\(A=\left(x-3y\right)\left(2x+y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
d) \(2x^2-3x-27\)
\(=\left(2x^2+6x\right)-\left(9x+27\right)\)
\(=2x\left(x+3\right)-9\left(x+3\right)\)
\(=\left(2x-9\right)\left(x+3\right)\)
e) \(2x^2-5xy-3y^2\)
\(=\left(2x^2+xy\right)-\left(6xy+3y^2\right)\)
\(=2x\left(x+y\right)-3y\left(x+y\right)\)
\(=\left(2x-3y\right)\left(x+y\right)\)
Ta có :
2x^2-5xy-3y^2
= 2^x + xy - 6xy - 3y^2
= x(2x + y) - 3y(2x + y)
= (2x + y)(x - 3y)