Phân tích đa thức thành nhân tử :
a) \(x^3-x+3x^2y+3xy^2-y\)
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\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(x^3+y^3-3x^2+3x-1\\=(x^3-3x^2+3x-1)+y^3\\=(x-1)^3+y^3\\=(x-1+y)[(x-1)^2-(x-1)y+y^2]\\=(x+y-1)(x^2-2x+1-xy+y+y^2)\)
\(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)
x3 - x + 3x2y + 3xy2 + y3 - y ( sửa -x3 -> x3 )
= ( x3 + 3x2y + 3xy2 + y3 ) - ( x + y )
= ( x + y )3 - ( x + y )
= ( x + y )[ ( x + y )2 - 1 ]
= ( x + y )( x + y - 1 )( x + y + 1 )
a) \(=2\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
\(=2\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left(2-x+y\right)\)
b) \(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+y^3\right)+\left(3x^2+3xy^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+3xy-1\right)\)
\(=\left(x+y\right)\left(x^2+y^2+2xy-1\right)\)
x³ - 3x²y + 3xy² - y³ - z³
= (x³ - 3x²y + 3xy² - y³) - z³
= (x - y)³ - z³
= (x - y - z)[(x - y)² + (x - y)z + z²]
= (x - y - z)(x² - 2xy + y² + xz - yz + z³)
--------------------
x² - y² + 8x + 6y + 7
= (x² + 8x + 16) - (y² - 6y + 9)
= (x + 4)² - (y - 3)²
= (x + 4 - y + 3)(x + 4 + y - 3)
= (x - y + 7)(x + y + 1)
a: \(=\left(x^3-3x^2y+3xy^2-y^3\right)-z^3\)
\(=\left(x-y\right)^3-z^3\)
\(=\left(x-y-z\right)\left[\left(x-y\right)^2+z\left(x-y\right)+z^2\right]\)
\(=\left(x-y-z\right)\left(x^2-2xy+y^2+xz-yz+z^2\right)\)
b: \(=x^2+8x+16-y^2+6y-9\)
=(x+4)^2-(y-3)^2
=(x+4+y-3)(x+4-y+3)
=(x+y+1)(x-y+7)
\(x^3-3x^2y+3xy^2-y^3-z^3\)
\(=\left(x-y\right)^3-z^3\)
\(=\left(x-y-z\right)\left[\left(x-y\right)^2+z\left(x-y\right)+z^2\right]\)
\(x^3-3x^2y+3xy^2-y^3-z^3\\ =\left(x-y\right)^3-z^3\\ =\left(x-y-z\right)\left(x^2-2xy+y^2+zy-xz+z^2\right)\\ =\left(x-y-z\right)\left(x^2+y^2+z^2+2xy+zy-zx\right)\)
Thiếu y3 nha bạn :
\(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)