Phân tích đa thức thành nhân tử:
x3 - 6x2 + x2y + 9x - 3y
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a: \(6x^2-3x\)
\(=3x\cdot2x-3x\)
=3x(2x-1)
b: \(15x^5y^4+10x^3y^3-5xy\)
\(=5xy\cdot3x^4y^3+5xy\cdot2x^2y^2-5xy\cdot1\)
\(=5xy\left(3x^4y^3+2x^2y^2-1\right)\)
c: \(x^2y+4xy+4y\)
\(=y\cdot x^2+y\cdot4x+y\cdot4\)
\(=y\left(x^2+4x+4\right)=y\left(x+2\right)^2\)
\(x^3-y^3+2x^2+2xy\)
\(=x\left(x^2-y^2+2x+2y\right)\)
\(=\)\(x\left[\left(x+y\right)\left(x-y\right)+2\left(x+y\right)\right]\)
\(=x\left(x+y\right)\left(x-y+2\right)\)
\(=x^3+2+3x^3-6=4x^3-4=4\left(x^3-1\right)=4\left(x-1\right)\left(x^2+x+1\right)\)
\(a,=5xy\left(2x-y+3z\right)\\ b,=x^2\left(x-1\right)-4\left(x-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)\\ c,=x\left(x^2-6x+9\right)=x\left(x-3\right)^2\)
Câu 1:
$x^2+4y^2+4xy-16=[x^2+(2y)^2+2.x.2y]-16$
$=(x+2y)^2-4^2=(x+2y-4)(x+2y+4)$
Câu 2:
$x^3+x^2+y^3+xy=(x^3+y^3)+(x^2+xy)$
$=(x+y)(x^2-xy+y^2)+x(x+y)=(x+y)(x^2-xy+y^2+x)$
Câu 1:
\(x^2+4y^2+4xy-16\)
\(=\left(x+2y\right)^2-16\)
\(=\left(x+2y+4\right)\left(x+2y-4\right)\)
Câu 2:
\(x^3+x^2+y^3+xy\)
\(=\left(x^3+y^3\right)\left(x^2+xy\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+x\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+x\right)\)
Ta có: \(x^3-\left(y-2\right)^3+\left(y-x-2\right)^2\)
\(=\left(x-y+2\right)\left(x^2+xy-2x+y^2-4y+4\right)+\left(x-y+2\right)^2\)
\(=\left(x-y+2\right)\left(x^2+xy-2x+y^2-4y+4+x-y+2\right)\)
\(=\left(x-y+2\right)\left(x^2+y^2+6+xy-x-5y\right)\)
\(a,=ab\left(a+3\right)\\ b,=\left(x-1\right)^2\\ c,=x\left[\left(x-3\right)^2-y^2\right]=x\left(x-y-3\right)\left(x+y-3\right)\)
3) \(x^2\left(x+2y\right)-x-2y\)
\(=x^2\left(x+2y\right)-\left(x+2y\right)\)
\(=\left(x^2-1\right)\left(x+2y\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(x+2y\right)\)
4) \(x^3-4x^2-9x+36\)
\(=\left(x^3-4x^2\right)-\left(9x-36\right)\)
\(=x^2\cdot\left(x-4\right)-9\left(x-4\right)\)
\(=\left(x-4\right)\left(x^2-9\right)\)
\(=\left(x-4\right)\left(x+3\right)\left(x-3\right)\)
\(x^2\left(x+2y\right)-x-2y\\ =x^2\left(x+2y\right)-\left(x+2y\right)\\ =\left(x^2-1\right)\left(x+2y\right)\\ =\left(x-1\right)\left(x+1\right)\left(x+2y\right)\\ ---\\ x^3-4x^2-9x+36\\ =x^2\left(x-4\right)-9\left(x-4\right)\\ =\left(x^2-9\right)\left(x-4\right)\\ =\left(x-3\right)\left(x+3\right)\left(x-4\right)\)
Ta có
\(x^3-6x^2+x^2y+9x-3y\\ =\left(x^3-6x^2+9x\right)+\left(x^2y-3y\right)\\ =x\left(x^2-3\right)^2+y\left(x^2-3\right)\)
=(x^2-3)(x+y)