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Bài 1:
a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)
= \(x^2\) - 16 - 5\(x\) - 5 + \(x^2\) + \(x\)
= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)
= 2\(x^2\) - 4\(x\) - 21
b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)
= 3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7
= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)
= - 3\(x^2\) + 3\(xy\) - 3
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left(3x+2\right)\left(9x^2-6x+4\right)\)
\(=27x^3+8\)
b: \(\left(x-2y\right)^3-\left(x^2-2xy+y^2\right)\)
\(=x^3-6x^2y+12xy^2-8y^3-x^2+2xy-y^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm
\(\frac{4x^3y^2-6x^2y^3}{2xy+2xy\left(y-x\right)}=\frac{2x^2y^2\left(2x-3y\right)}{2xy\left(1+y-x\right)}=\frac{xy\left(2x-3y\right)}{1+y-x}\)
Học tốt
\(\frac{4x^3y^2-6x^2y^3}{2xy+2xy\left(y-x\right)}=\frac{2x^2y^2\left(2x-3y\right)}{2xy\left(1+y-x\right)}=\frac{xy\left(2x-3y\right)}{y-x+1}\)
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a: =xy(1/3+4-2)=7/3xy
b: =xy^2(-1+3/2+4/3)=(1/3+3/2)xy^2=11/6xy^2
c: =4x^2y^2+2/3x^2y^2-4/3x^2y=-4/3x^2y+14/3x^2y^2
d: =3x^2y^2z+4x^2y^2z-8x^2y^2z=-x^2y^2z
![](https://rs.olm.vn/images/avt/0.png?1311)
B) Ta có: 2x-2y-x2+2xy-y2
⇔ 2(x-y)-(x2-2xy+y2)
⇔ 2(x-y)-(x-y)2
⇔ (x-y)(2-x+y)
Đúng thì tick nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
\(\Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
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1. \(\dfrac{x^3-4x^2+4x}{x^2-4}=\dfrac{x\left(x^2-4x+4\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)}{x+2}\)
\(\dfrac{2xy-x^2}{3x^3-6x^2y}\\ =\dfrac{x\left(2y-x\right)}{3x^2\left(x-2y\right)}\\ =\dfrac{x\left(2y-x\right)}{-3x^2\left(2y-x\right)}\\ =\dfrac{1}{-3x}\)
\(\dfrac{2xy-x^2}{3x^3-6x^2y}=\dfrac{-\left(x^2-2xy\right)}{3x^3-6x^2y}\)
\(=\dfrac{-x\left(x-2y\right)}{3x^2\left(x-2y\right)}=\dfrac{-1}{3x}\)