Phân tích đa thức thành nhân tử :
a ) x ^ 3 - 2x ^ 2 - 9xy ^ 2 + x
b ) ( x +2 ) ( x + 3 ) ( x +4 ) (x + 5 ) - 8
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KN
9 tháng 9 2019
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) \(x^3+3x^2y-9xy^2+5y^3\)
\(=x^3+5x^2y-2x^2y-10xy^2+xy^2+5y^3\)
\(=\left(5y^3-10xy^2+5x^2y\right)+\left(xy^2-2x^2y+x^3\right)\)
\(=5y\left(y^2-2xy+x^2\right)+x\left(y^2-2xy+x^2\right)\)
\(=\left(5y+x\right)\left(y^2-2xy+x^2\right)\)
\(=\left(5y+x\right)\left(y-x\right)^2\)
14 tháng 8 2021
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: \(x^3+9x^2-4x-36\)
\(=x^2\left(x+9\right)-4\left(x+9\right)\)
\(=\left(x+9\right)\left(x-2\right)\left(x+2\right)\)
14 tháng 8 2021
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) \(x^3+9x^2-4x-36\\ =\left(x^3-2x^2\right)+\left(11x^2-22x\right)+\left(18x-36\right)\\ =x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\\ =\left(x^2+11x+18\right)\left(x-2\right)\\ =\left[\left(x^2+2x\right)+\left(9x+18\right)\right]\left(x-2\right)\\ =\left[x\left(x+2\right)+9\left(x+2\right)\right]\left(x-2\right)\\ =\left(x+2\right)\left(x+9\right)\left(x-2\right)\)
29 tháng 8 2018
mk viết đáp án, ko biết biến đổi ib mk
a) \(x^3+3x^2y-9xy^2+5y^3=\left(x+5y\right)\left(x-y\right)^2\)
b) \(x^4+x^3+6x^2+5x+5=\left(x^2+5\right)\left(x^2+x+1\right)\)
c) \(x^4-2x^3-12x^2+12x+36=\left(x^2-6\right)\left(x^2-2x-6\right)\)
d) \(x^8y^8+x^4y^4+1=\left(x^2y^2-xy+1\right)\left(x^2y^2+xy+1\right)\left(x^4y^4-x^2y^2+1\right)\)
TP
2
15 tháng 8 2016
a) x3 +x+2
=\(\left(x^3+x^2\right)-\left(x^2+x\right)+\left(2x+2\right)\)
=\(\left(x+1\right)\left(x^2-x+2\right)\)
b) x3-2x-1
=\(\left(x^3+x^2\right)-\left(x^2+x\right)-\left(x+1\right)\)
=\(\left(x+1\right)\left(x^2-x-1\right)\)
c) x3+3x2-4
=\(\left(x^3-x^2\right)+\left(4x^2+4x\right)-\left(4x+4\right)\)
=\(\left(x-1\right)\cdot\left(x^2+4x-4\right)\)
d) x3+3x2y-9xy2+5y3
=\(\left(x^3-x^2y\right)+\left(4x^2y-4xy^2\right)-\left(5xy^2-5y^3\right)\)
=\(\left(x-y\right)\left(x^2+4xy-5y^2\right)\)
=\(\left(x-y\right)^2\left(x-5y\right)\)
IM
15 tháng 8 2016
a)
\(x^3+x+2\)
\(=\left(x^3+x^2\right)-\left(x^2+x\right)+\left(2x+2\right)\)
\(=x^2\left(x+1\right)-x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+2\right)\)
b)
\(x^3-2x-1\)
\(=\left(x^3+x^2\right)-\left(x^2+x\right)-\left(x+1\right)\)
\(=x^2\left(x+1\right)-x\left(x+1\right)-\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-1\right)\)
c)
\(x^3-3x^2-4\)
\(=\left(x^3-x^2\right)+\left(4x^2-4x\right)+\left(4x-4\right)\)
\(=x^2\left(x-1\right)+4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+2.2.x+2^2\right)\)
\(=\left(x-1\right)\left(x+2\right)^2\)
d)
\(x^3-3x^2y-9xy^2+5y^3\)
\(=\left(x^3-x^2y\right)+\left(4x^2y-4xy^2\right)-\left(5xy^2-5y^3\right)\)
\(=x^2\left(x-y\right)+4xy\left(x-y\right)-5y^2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-4xy-5y^2\right)\)
\(=\left(x-y\right)^2\left(x-5y\right)\)
