Phân tích đa thức thành nhân tử: x^2+xy+y^2-x^2y^2
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\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
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\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
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\(x^3-x^{22}-xy^2+y^2\)
\(=x^2\left(x-1\right)-y^2\left(x-1\right)\)
\(=\left(x^2-y^2\right)\left(x-1\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x-1\right)\)
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1) \(2\left(x-1\right)^3-\left(x-1\right)=\left(x-1\right)\left(2\left(x-1\right)^2-1\right)\)
2) \(y\left(x-2y\right)^2+xy^2\left(2y-x\right)=\left(2y-x\right)\left(2\left(2y-x\right)+1\right)=\left(2y-x\right)\left(4y-2x+1\right)\)
3) \(xy\left(x+y\right)-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\) (xem lại đề sửa -2x thành -x mới đúng)
4) \(xy\left(x-3y\right)-2x+6y=xy\left(x-3y\right)-2\left(x-3y\right)=\left(x-3y\right)\left(xy-2\right)\)
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Ta có: x2 + y2 - x2y2 + xy - x - y
= (x2 - x2y2) + (y2 - y) + (xy - x)
= - x2(y2 - 1) + y(y - 1) + x(y - 1)
= - x2(y + 1)(y - 1) + (y - 1)(x + y)
= (y - 1)(x + y - x2y - x2)
= (y - 1)[- (x2 - x) - (x2y - y)]
= - (y - 1)[x(x - 1) + y(x2 - 1)]
= - (y - 1)[x(x - 1) + y(x + 1)(x - 1)]
= - (y - 1)(x - 1)[x + y(x + 1)]
= - (y - 1)(x - 1)(x + xy +y)
Ta có: x2 + y2 - x2y2 + xy - x - y
= (x2 - x2y2) + (y2 - y) + (xy - x)
= - x2(y2 - 1) + y(y - 1) + x(y - 1)
= - x2(y + 1)(y - 1) + (y - 1)(x + y)
= (y - 1)(x + y - x2y - x2)
= (y - 1)[- (x2 - x) - (x2y - y)]
= - (y - 1)[x(x - 1) + y(x2 - 1)]
= - (y - 1)[x(x - 1) + y(x + 1)(x - 1)]
= - (y - 1)(x - 1)[x + y(x + 1)]
= - (y - 1)(x - 1)(x + xy +y)
Ai trên 10 điểm hỏi đáp thì mình nha mình đang cần gấp chỉ còn 59 điểm là tròn rồi mong các bạn hỗ trợ mình sẽ đền bù xứng đáng
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\(x\left(x-y\right)^2-y\left(x-y\right)^2+xy^2-x^2y\)
\(=\left(x-y\right)^2\left(x-y\right)-xy\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-2xy+y^2-xy\right)\)
\(=\left(x-y\right)\left(x^2-3xy+y^2\right)\)