Phân tích thành nhân tử:
6x^2y + 9xy^2 -2 - 3y
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1: \(6x^2y-9xy^2+3xy\)
\(=3xy\left(2x-3y+1\right)\)
2: \(\left(4-x\right)^2-16\)
\(=\left(4-x-4\right)\left(4-x+4\right)\)
\(=-x\cdot\left(8-x\right)\)
3: \(x^3+9x^2-4x-36\)
\(=x^2\left(x+9\right)-4\left(x+9\right)\)
\(=\left(x+9\right)\left(x-2\right)\left(x+2\right)\)
1) \(6x^2y-9xy^2+3xy=3xy\left(2x-3y+1\right)\)
2) \(\left(4-x\right)^2-16=\left(4-x\right)^2-4^2=\left(4-x-4\right)\left(4-x+4\right)=-x\left(8-x\right)\)
3) \(x^3+9x^2-4x-36\\ =\left(x^3-2x^2\right)+\left(11x^2-22x\right)+\left(18x-36\right)\\ =x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\\ =\left(x^2+11x+18\right)\left(x-2\right)\\ =\left[\left(x^2+2x\right)+\left(9x+18\right)\right]\left(x-2\right)\\ =\left[x\left(x+2\right)+9\left(x+2\right)\right]\left(x-2\right)\\ =\left(x+2\right)\left(x+9\right)\left(x-2\right)\)
a) \(x^3y^3+x^2y^2+4\)
\(=x^3y^3-x^2y^2+2x^2y^2-2xy+2xy+4\)
\(=\left(x^3y^3-x^2y^2+2xy\right)+\left(2x^2y^2-2xy+4\right)\)
\(=xy\left(x^2y^2-xy+2\right)+2\left(x^2y^2-xy+2\right)\)
\(=\left(xy+2\right)\left(x^2y^2-xy+2\right)\)
b) \(x^3+3x^2y-9xy^2+5y^3\)
\(=x^3+5x^2y-2x^2y-10xy^2+xy^2+5y^3\)
\(=\left(5y^3-10xy^2+5x^2y\right)+\left(xy^2-2x^2y+x^3\right)\)
\(=5y\left(y^2-2xy+x^2\right)+x\left(y^2-2xy+x^2\right)\)
\(=\left(5y+x\right)\left(y^2-2xy+x^2\right)\)
\(=\left(5y+x\right)\left(y-x\right)^2\)
phân tích đa thức thành nhân tử
a, 6x^2 + 7xy + 2y^2
=6x^2+3xy+4xy+2y^2
=3x(x+y)+2y(x+y)
=(3x+2y)(x+y)
b, 9x^2 - 9xy - 4y^2
=9x^2 +3xy-12xy-4y^2
=3x(x+y)-4y(x+y)
=(3x+4y)(x+y)
c, x^2 - y^2 + 10x - 6y + 16=x^2-y^2+6x-6y+4x+16=x(x+6)-y(x+6)+4(x+6)=(x-y+4)(x+6)
Bài làm
a, 6x2 + 7xy + 2y2
= 6x2 + 3xy + 4xy + 2y2
= ( 6x2 + 3xy ) + ( 4xy + 2y2 )
= 3x( 2x + y ) + 2y( 2x + y )
= ( 2x + y )( 3x + 2y )
b, 9x2 - 9xy - 4y2
= 9x2 - 12xy + 3xy - 4y2
= ( 9x2 - 12xy ) + ( 3xy - 4y2 )
= 3x( 3x - 4y ) + y ( 3x - 4y )
= ( 3x + y )( 3x - 4y )
c, x2 - y2 + 10x - 6y + 16
= x2 - y2 - 6x + 6y + 4x + 16
= x( x + 6 ) - y( x + 6 ) + 4( x + 6 )
= ( x - y + 4 )( x + 6 )
# Học tốt #
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) \(x^3+2x^2-xy-2y\)
\(=\left(x^3-xy\right)+\left(2x^2-2y\right)\)
\(=x\left(x^2-y\right)+2\left(x^2-y\right)\)
\(=\left(x+2\right)\left(x^2-y\right)\)
\(=\left(x+2\right)\left(x+\sqrt{y}\right)\left(x-\sqrt{y}\right)\)
\(3x^3y-6x^2y-3xy^3-6xy^2z-3xyz^2+3xy\)
\(=3xy\left(x^2-2x-y^2-2yz-z^2+1\right)\)
\(=3xy\left[\left(x-1\right)^2-\left(y+z\right)^2\right]\)
\(=3xy\left(x-1-y-z\right)\left(x-1+y+z\right)\)
đề bạn thế này à:
\(6x^2y+9xy^2-2-3y\)