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`a, -xy(x^2+xy-y^2)`
`= -x^3y - x^2y^2 + xy^3`.
`b, 5x^2y(2y^2-xy)`
`= 10x^2y^3 - 5x^3y^2`.
`c, (-2x^3 - 1/4y - 4y^2).8xy^2`.
`= -16x^4y^2 - 2xy^3 - 32xy^4`.
`d, (2x^3 - 3xy + 12x)(-1/6xy)`
`= -2/3x^4y + 1/2x^2y^2 - 2x^2y`.
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a: \(=\dfrac{3x+6-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+10}{x^2-4}\)
b: \(=\dfrac{10x+15+4x-6+2x+5}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{16x+14}{\left(2x-3\right)\left(2x+3\right)}\)
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a) (x-2)(x+2)-x(x-1)+8
= x2-4-x2+x+8
= (x2-x2)+(-4+8)+x
= 4+x
b) bn viết lại đề đi:v
đọc khó quá.
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a,
$xy^2+x^2y+(-2xy^2)=xy^2-2xy^2+x^2y=-xy^2+x^2y$
b,
$12x^2y^3z^4+(-7x^2y^3z^4)=12x^2y^3z^4-7x^2y^3z^4=5x^2y^3z^4$
c,
$-6xy^3-(-6xy^3)+6x^3=-6xy^3+6xy^3+6x^3=0+6x^3=6x^3$
d,
$\frac{-x^2}{2}+\frac{7}{2}x^2+x=(\frac{7}{2}-\frac{1}{2})x^2+x$
$=3x^2+x$
e,
$2x^3+3x^3-\frac{1}{3}x^3=(2+3-\frac{1}{3})x^3=\frac{14}{3}x^3$
f,
$5xy^2+\frac{1}{2}xy^2+\frac{1}{4}xy^2=(5+\frac{1}{2}+\frac{1}{4})xy^2$
$=\frac{23}{4}xy^2$
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a) ( 3x + 2y - 1 )( x - 5 ) - ( x - 2 )2y
= 3x(x - 5) + 2y(x - 5) - 1(x - 5) - ( 2xy - 4y )
= 3x2 - 15x + 2xy - 10y - x + 5 - 2xy + 4y
= 3x2 - 16x - 6y + 5
b) ( 3x - 2 )( 3x + 2 ) - ( 2x + 1 )( 4x + 3 )
= [ ( 3x )2 - 22 ] - ( 8x2 + 10x + 3 )
= 9x2 - 4 - 8x2 - 10x - 3
= x2 - 10 - 7
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a: =1/3x^3-2x^2-7/3x
b: =5/4xyz^2+1/4xyz
c: =9/8x^2y^3-3x^2y
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người dễ theo dõi hơn.
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a: \(=\dfrac{5}{3}x^2-x+\dfrac{1}{3}\)
b: \(=-5y-9+xy\)
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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
(2x2y-2y+3).x2.(-1/2y)
= -x4y2+x2 y2 -3/2 x2y
HT