Phân tích đa thức thành nhân tử ( kết hợp nhiều pp ) :
a, x\(^2\) - xz - 9y\(^2\) + 3yz
b, 12x\(^3\) + 4x\(^2\)\(^{ }\) - 27xy - 9y
c, x\(^4\) - 25x\(^2\) + 20x - 4
d, x\(^2\)y\(^2\).( x\(^2\) - 9 ) - x\(^2\) + 9
Phân tích đa thức thành nhân tử :
1, 4 - x\(^2\) - 2xy - y\(^2\)
2, 25 - x\(^2\) + 2xy - y\(^2\)
3, 4x\(^2\) + 4xy + y\(^2\) - 9
4, x\(^2\) + 4xy - 16 + 4y\(^2\)
a) \(x^2-xz-9y^2+3yz=\left(x^2-9y^2\right)-z\left(x-3y\right)\)
\(=\left(x+3y\right)\left(x-3y\right)-z\left(x-3y\right)=\left(x-3y\right)\left(x+3y-z\right)\)
b) \(12x^3+4x^2-27xy-9y=4x^2\left(3x+1\right)-9y\left(3x+1\right)=\left(3x+1\right)\left(4x^2-9y\right)\)
c) \(x^4-25x^2+20x-4=x^4-\left(25x^2-20x+4\right)=x^4-\left(5x-2\right)^2=\left(x^2-5x+2\right)\left(x^2+5x-2\right)\)d) \(x^2y^2\cdot\left(x^2-9\right)-x^2+9=x^2y^2\cdot\left(x^2-9\right)-\left(x^2-9\right)=\left(x^2-9\right)\left(x^2y^2-1\right)\)
\(=\left(x-3\right)\left(x+3\right)\left(xy-1\right)\left(xy+1\right)\)
1) \(4-x^2-2xy-y^2=4-\left(x^2+2xy+y^2\right)=4-\left(x+y\right)^2=\left(2-x-y\right)\left(2+x+y\right)\)
2) \(25-x^2+2xy-y^2=25-\left(x^2-2xy+y^2\right)=25-\left(x-y\right)^2=\left(5-x+y\right)\left(5+x-y\right)\)3) \(4x^2+4xy+y^2-9=\left(4x^2+4xy+y^2\right)-9=\left(2x+y\right)^2-9=\left(2x+y-3\right)\left(2x+y+3\right)\)4)\(x^2+4xy-16+4y^2=\left(x^2+4xy+4y^2\right)-16=\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\)
\(1)4-x^2-2xy-y^2 =4-(x^2+2xy+y^2) =4-(x+y)^2 \)
\(2)25-x^2+2xy-y^2 =5^2-(x^2-2xy+y^2) =5^2-(x-y)^2 =(5-x-y)(5+x-y)\)
\(3)4x^2+4xy+y^2-9 =(2x)^2+4xy+y^2-3^2 =(2x+y)^2-3^2 =(2x+y-3)(2x+y+3)\)
\(4)x^2+4xy-16+4y^2 =x^2+4xy+4y^2-16 =(x-4y)^2-4^2 =(x-4y-4)(x-4y+4)\)
\(a)x^2-xz-9y^2+3yz =(x^2-9y^2)-(xz-3yz) =(x+3y)(x+3y)-z(x-3y) =(x-3y)(x+3y-z)\)
\(b)12x^3+4x^2-27xy-9y =4x^2(3x+1)-9y(3x+1) =(3x+1)(4x^2-9y)\)
\(c)x^4-25x^2+20x-4 =x^4-25x^2+20x-2^2 =x^4-((5x)^2-20x+2^2) =(x^2)^2-(5x-2)^2 =(x^2-5x-2)(x^2+5x-2)\)
\(d)x^2y^2.(x^2-9)-x^2+9 =x^2y^2.(x^2-9)-(x^2-y) =(x^2-9)(x^2y^2-1)\)