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\(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
\(x^3-x+3x^2+3xy^2+y^3-y\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)
a) x2-x=x(x-1)
b) 3x-3y = 3(x-y)
c)3xy2+6xyz=3xy(y+2z)
d) 5x-20y=5(x-4y)
g) 5x(x-1)-(1-x)=5x(x-1)+(x-1)=(5x+1)(x-1)
a, \(x^2-x=x\left(x-1\right)\)
b, \(3x-3y=3\left(x-y\right)\)
c, \(3xy^2+6xyz=3xy\left(y+2z\right)\)
d, \(5x-20y=5\left(x-4y\right)\)
g, \(5x\left(x-1\right)-\left(1-x\right)=\left(5x+1\right)\left(x-1\right)\)
\(a,=3xy\left(x-2y\right)\\ b,=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)=\left(x+y+3\right)\left(x-y\right)\\ c,=x\left[\left(x+2\right)^2-y^2\right]=x\left(x+y+2\right)\left(x-y+2\right)\\ d,\Leftrightarrow x\left(x^2-4\right)=0\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Lời giải:
Ta có:
\(3x^3-6x^2y+3xy^2-3x^2=3x(x^2-2xy+y^2-x)\)
a. 3xy( 4x + y - \(\dfrac{4}{3}\) )
b. 2x2( 3x + 1 )
c. (2x + 3 )( x - y )
d. xy( 1 - x )( x - 1 )
e. 6( 2x + 1 )( x + y )
Phân tích đa thức này thành nhân tử.
x3−3x2y+3xy2−y3+y2−x2
e: \(x^2+6x+9-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3-y\right)\left(x+3+y\right)\)
f: \(x^2-2x+7x-14\)
\(=x\left(x-2\right)+7\left(x-2\right)\)
=(x-2)(x+7)
h: \(5x^2-10xy+5y^2-20\)
\(=5\left(x^2-2xy+y^2-4\right)\)
\(=5\left(x-y-2\right)\left(x-y+2\right)\)
a: \(3x^4-6x^3+2x^2=x^2\left(3x^2-6x+2\right)\)
b: \(x^3y+12x^2y+36xy=xy\left(x^2+12x+36\right)=xy\left(x+6\right)^2\)
c: \(x^3y-9xy^3=xy\left(x^2-9y^2\right)=xy\left(x-3y\right)\left(x+3y\right)\)
d: \(x^2y^2-2xy^2+y^2=y^2\left(x-1\right)^2\)