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\(x^3-3xy^2-2y^3\)
\(=x^3-xy^2-2xy^2-2y^3\)
\(=x\left(x^2-y^2\right)-2y^2\left(x+y\right)\)
\(=x\left(x-y\right)\left(x+y\right)-2y^2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy-2y^2\right)\)
\(=\left(x+y\right)\left(x^2-2xy+xy-2y^2\right)\)
\(=\left(x+y\right)\left[x\left(x-2y\right)+y\left(x-2y\right)\right]\)
\(=\left(x+y\right)^2\left(x-2y\right)\)
\(x^3-3xy^2-2y^3\)
\(=x^3-xy^2-2xy^2-2y^3\)
\(=x\left(x^2-y^2\right)-2y^2\left(x+y\right)\)
\(=x\left(x-y\right)\left(x+y\right)-2y^2\left(x+y\right)\)
\(=\left(x+y\right)\left[x\left(x+y\right)-2y^2\right]\)
\(=\left(x+y\right)\left(x^2+xy-2y^2\right)\)
\(=\left(x+y\right)\left(x^2+2xy-xy-2y^2\right)\)
\(=\left(x+y\right)\left[x\left(x-2y\right)-y\left(x-2y\right)\right]\)
\(=\left(x-y\right)\left(x-y\right)\left(x-2y\right)\)
\(=\left(x-y\right)^2\left(x-2y\right)\)
Thiếu y3 nha bạn :
\(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
Làm tính nhân
(4x3+3xy2-2y3).(3x2-5xy-6y2)
=12x5+12y5-20x4y-36x2y3-8xy4
Phân tích đa thức thành nhân tử
10x3+5x2y-10x2y-10xy2+5y3
=10x3-5x2y-10xy2+5y3
=5(2x3-x2y-2xy2+y3-)
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) x2 - xy - 20y2
= x2 + 4xy - 5xy - 20y2
= x( x + 4y ) - 5y( x + 4y )
= ( x + 4y )( x - 5y )
b) x3 - x2y - 3xy2 + 2y3
= x3 + x2y - 2x2y - xy2 - 2xy2 + 2y3
= ( x3 + x2y - xy2 ) - ( 2x2y + 2xy2 - 2y3 )
= x( x2 + xy - y2 ) - 2y( x2 + xy - y2 )
= ( x2 + xy - y2 )( x - 2y )
\(5x^3-5x=5x\left(x^2-1\right)\)
\(3x^2+5x-3xy-5x=x\left(3x+5\right)-x\left(3y+5\right)=x\left(3x-3y\right)=3x\left(x-y\right)\)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
1/
x2 - 3x - 4
= \(x^2-3x+\frac{9}{4}-\frac{9}{4}-4\)
\(=\left(x^2-3x+\frac{9}{4}\right)-\frac{25}{4}\)
\(=\left(x-\frac{3}{2}\right)^2-\left(\frac{5}{2}\right)^2\)
\(=\left(x-\frac{3}{2}-\frac{5}{2}\right)\left(x-\frac{3}{2}+\frac{5}{2}\right)\)
\(=\left(x-4\right)\left(x+1\right)\)
Bài 1 :
\(x^2-3x-4\)
\(=x^2+x-4x-4\)
\(=x\left(x+1\right)-4\left(x+1\right)\)
\(=\left(x+1\right)\left(x-4\right)\)