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a) \(2xy-y+6x-3=\left(2xy+6x\right)-\left(y+3\right)=2x\left(y+3\right)-\left(y+3\right)=\left(2x-1\right)\left(y+3\right)\)
b) \(x^2-2xy-x+2y=\left(x^2-2xy\right)-\left(x-2y\right)=x\left(x-2y\right)-\left(x-2y\right)=\left(x-1\right)\left(x-2y\right)\)
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) \(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+y^3\right)+\left(3x^2+3xy^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+3xy-1\right)\)
\(=\left(x+y\right)\left(x^2+y^2+2xy-1\right)\)
Lời giải:
$y-x^2y-2xy^2-y^3=y(1-x^2-2xy-y^2)$
$=y[1-(x^2+2xy+y^2)]=y[1-(x+y)^2]=y(1-x-y)(1+x+y)$
a) Ta có: \(x^2y^2-x^2+6xy-9y^2\)
\(=x^2y^2-\left(x^2-6xy+y^2\right)\)
\(=\left(xy\right)^2-\left(x-3y\right)^2\)
\(=\left(xy-x+3y\right)\left(xy+x-3y\right)\)
b) Ta có: \(9-x^2+2xy-y^2\)
\(=9-\left(x^2-2xy+y^2\right)\)
\(=9-\left(x-y\right)^2\)
\(=\left(9-x+y\right)\left(9+x-y\right)\)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
x2 + y2 + 2xy - 2x - 2y
= (x2 + 2xy + y2) - (2x + 2y)
= (x + y)2 - 2(x + y)
= (x + y)(x + y + 2)
\(x^2+y^2-2x-2y+2xy-3\)
\(=x^2+y^2+1-2x-2y+2xy-4\)
\(=\left(x+y-1\right)^2-2^2\)
\(=\left(x+y-3\right).\left(x+y+1\right)\)
\(=y\left(1-x^2-2xy-y^2\right)=y\left[1-\left(x+y\right)^2\right]=y\left(1-x-y\right)\left(1+y+x\right)\)