Phân tích đa thức thành nhân tử:
\(x^2y\) + \(x^2\) - \(y\) - 1
Các bạn giúp mik vs ạ1
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\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(2xy-x^2-y^2+16\)
\(=-\left(x^2-2xy+y^2-16\right)\)
\(=-\left(x-y-4\right)\left(x-y+4\right)\)
b) \(5x^3+10x^2y+5xy^2=2\left(x^3+2x^2y+xy^2\right)\)
\(=2\left(x^3+x^2y+x^2y+xy^2\right)=2\left[x^2\left(x+y\right)+xy\left(x+y\right)\right]\)
=\(2\left(x^2+xy\right)\left(x+y\right)\)
h) \(=3x\left(2y-3z\right)\left[x^2-5\left(2y-3z\right)\right]=3x\left(2y-3z\right)\left(x^2-10y+15z\right)\)
k) \(=\left(x+2\right)\left(3x-5\right)\)
l) \(=\left(18^2+3\right)\left(x+3\right)=327\left(x+3\right)\)
m) \(=7xy\left(2x-3y+4xy\right)\)
n) \(=2\left(x-y\right)\left(5x-4y\right)\)
\(x\left(x-y\right)+2\left(y-x\right)=x\left(x-y\right)-2\left(x-y\right)=\left(x-y\right)\left(x-2\right)\)
\(=x\left(x-y\right)-2\left(x-y\right)=\left(x-2\right)\left(x-y\right)\)
\(=x^2\left(y+1\right)-\left(y+1\right)\)
=(y+1)(x-1)(x+1)