phân tích đa thức 6x^2-xy-4y^2 thành nhân tử
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\(=\left(x^2-6x+9\right)-4y^2\)
\(=\left(x-3\right)^2-\left(2y\right)^2\)
\(=\left(x-3-2y\right)\left(x-3+2y\right)\)
= ( x^2 - 4y^2 ) + ( 9 - 6x)
= [ x^2 - (2y)^2 ] + 3( 3 - 2x )
= (x - 2y)(x + 2y)+ 3(3 - 2x)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. \(x^3+2x^2-6x-27=\left(x-3\right)\left(x^2+5x+9\right)\)
2. \(9x^2+6x-4y^2-4y=\left(9x^2-4y^2\right)+\left(6x-4y\right)\)
\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)=\left(3x-2y\right)\left(3x+2y+2\right)\)
3. \(12x^3+4x^2-27x-9=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(x^2-\dfrac{9}{4}\right)=\left(x+\dfrac{1}{3}\right)\left(x+\dfrac{3}{2}\right)\left(x-\dfrac{3}{2}\right)\)
1) Ta có: \(x^3+2x^2-6x-27\)
\(=\left(x-3\right)\left(x^2+3x+9\right)+2x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
2: Ta có: \(9x^2+6x-4y^2-4y\)
\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(3x+2y+2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-\left(x+2y\right)^2\)
\(-\left(x-3\right)^2\)
\(\left(3-5x\right)^2\)
\(-x^2-4xy-4y^2=-\left(x+2y\right)^2\)
\(-x^2+6x-9=-\left(x-3\right)^2\)
\(25x^2-30x+9=\left(5x-3\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
6x^2-xy-7y^2
=( - ( y + x )) ( 7y - 6x )
nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=x\left(x-6\right)+y\left(x-6\right)=\left(x+y\right)\left(x-6\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^4y-3x^3y^2+3x^2y^3+xy^4=xy\left(x^3-3x^2y+3xy^2+y^3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(6x^2-xy-4y^2=\left(6x^2-3xy\right)+\left(2xy-4y^2\right)\)
\(=3x\left(2x-y\right)+2y\left(2x-y\right)\)
\(=\left(2x-y\right)\left(3x+2y\right)\)