cho mik hỏi
hãy phân tik đa thức sau thành nhân tử :
x^3-2x^2-9xy^2+x
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) \(x^3+3x^2y-9xy^2+5y^3\)
\(=x^3+5x^2y-2x^2y-10xy^2+xy^2+5y^3\)
\(=\left(5y^3-10xy^2+5x^2y\right)+\left(xy^2-2x^2y+x^3\right)\)
\(=5y\left(y^2-2xy+x^2\right)+x\left(y^2-2xy+x^2\right)\)
\(=\left(5y+x\right)\left(y^2-2xy+x^2\right)\)
\(=\left(5y+x\right)\left(y-x\right)^2\)
x3 + 3x2y - 9xy2 + 5y3
= ( x3 - 3x2y + 3xy2 - y3 ) + ( 6y3 - 12xy2 + 6 x2y )
= ( x - y )3 + 6y ( x - y )2
= ( x - y )2 ( x + 5y )
1: \(3x\left(x-1\right)+7x^2\left(x-1\right)\)
\(=\left(x-1\right)\left(7x^2+3x\right)\)
=x(7x+3)(x-1)
2: \(9\left(x+5\right)^2-\left(x-7\right)^2\)
\(=\left(3x+15\right)^2-\left(x-7\right)^2\)
=(3x+15+x-7)(3x+15-x+7)
=(4x+8)(2x+22)
\(=4\left(x+2\right)\cdot2\cdot\left(x+11\right)=8\left(x+2\right)\left(x+11\right)\)
3: \(x^2-2xy+y^2-xz+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
=(x-y)(x-y-z)
4: \(\left(x^2+2x\right)^2-2x^2-4x-3\)
\(=\left(x^2+2x\right)^2-2\cdot\left(x^2+2x\right)-3\)
\(=\left(x^2+2x-3\right)\left(x^2+2x+1\right)=\left(x+3\right)\left(x-1\right)\left(x+1\right)^2\)
5: \(9x^2y^2+15x^2y-21xy^2\)
\(=3xy\cdot3xy+3xy\cdot5x-3xy\cdot7y\)
=3xy(3xy+5x-7y)
6: \(x^3-2x^2y+xy^2-9xy^4\)
\(=x\left(x^2-2xy+y^2-9y^4\right)\)
\(=x\left\lbrack\left(x-y\right)^2-\left(3y^2\right)^2\right\rbrack=x\left(x-y-3y^2\right)\left(x-y+3y^2\right)\)
7: \(x^2+x-12\)
\(=x^2+4x-3x-12\)
=x(x+4)-3(x+4)
=(x+4)(x-3)
8: (x+1)(x+3)(x+5)(x+7)+15
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
\(=\left(x^2+8x+7\right)^2+8\left(x^2+8x+7\right)+15\)
\(=\left(x^2+8x+7+3\right)\left(x^2+8x+7+5\right)\)
\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)=\left(x^2+8x+10\right)\left(x+2\right)\left(x+6\right)\)
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) \(x^3+9x^2-4x-36\\ =\left(x^3-2x^2\right)+\left(11x^2-22x\right)+\left(18x-36\right)\\ =x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\\ =\left(x^2+11x+18\right)\left(x-2\right)\\ =\left[\left(x^2+2x\right)+\left(9x+18\right)\right]\left(x-2\right)\\ =\left[x\left(x+2\right)+9\left(x+2\right)\right]\left(x-2\right)\\ =\left(x+2\right)\left(x+9\right)\left(x-2\right)\)
\(a,=x-x^2=x\left(1-x\right)\\ b,=x^2+3x-2x-6=\left(x+3\right)\left(x-2\right)\\ c,=x^2+2x+3x+6=\left(x+2\right)\left(x+3\right)\)
a) \(x^2-2x^2+x=-x^2+x=-x\left(x-1\right)\)
b) \(x^2+x-6=\left(x^2+3x\right)-\left(2x+6\right)=x\left(x+3\right)-2\left(x+3\right)=\left(x-2\right)\left(x+3\right)\)
c) \(x^2+5x+6=\left(x^2+2x\right)+\left(3x+6\right)=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
Ta có:\(x^3-2x^2-9xy^2+x\)
\(=x\left(x^2-2x-9y^2+1\right)=x\left(x^2-2x+1-9y^2\right)\)
\(=x\left[\left(x-1\right)^2-9y^2\right]=x\left(x-3y-1\right)\left(x-1+3y\right)\)