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huongkarry
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Tống Lê Kim Liên
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Yukihira Souma
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Phạm Hồ Thanh Quang
15 tháng 7 2017 lúc 7:46

a) ĐKXĐ: \(x;y\ne0,x\ne\frac{y}{2},y\ne\frac{x}{2}\)
\(\frac{y}{2x^2-xy}+\frac{4x}{y^2-2xy}=\frac{y}{x\left(2x-y\right)}-\frac{4x}{y\left(2x-y\right)}\)\(=\frac{y^2-4x^2}{xy\left(2x-y\right)}=\frac{\left(y-2x\right)\left(y+2x\right)}{xy\left(2x-y\right)}\)
\(=\frac{-\left(y+2x\right)}{xy}\)

b) ĐKXĐ: \(x\ne2;x\ne-2\)
\(\frac{1}{x+2}+\frac{3}{x^2-4}+\frac{x-14}{\left(x^2+4x+4\right)\left(x-2\right)}\)\(=\frac{1}{x+2}+\frac{3}{\left(x-2\right)\left(x+2\right)}+\frac{x-14}{\left(x+2\right)^2\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x+2\right)+3\left(x+2\right)+x-14}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\frac{\left(x^2+4x+4\right)-16}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{\left(x+2\right)^2-16}{\left(x+2\right)^2\left(x-2\right)}=\frac{\left(x+2-4\right)\left(x+2+4\right)}{\left(x+2\right)^2\left(x-2\right)}\)\(=\frac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\frac{x+6}{\left(x+2\right)^2}\)

Buddy
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『Kuroba ム Tsuki Ryoo...
22 tháng 7 2023 lúc 8:28

`a)`

`3x(2xy - 5x^2y)`

`= 3x*2xy + 3x* (-5x^2y)`

`= 6x^2y - 15x^3y`

`b)`

`2x^2y (xy - 4xy^2 + 7y)`

`= 2x^2y * xy + 2x^2y * (-4xy^2) + 2x^2y * 7y`

`= 2x^3y^2 - 8x^3y^3 + 14x^2y^2`

`c)`

`(-2/3xy^2 + 6yz^2)*(-1/2xy)`

`= (-2/3xy^2)*(-1/2xy) + 6yz^2 * (-1/2xy)`

`= 1/3x^2y^3 - 3xy^2z^2`

`a, 3x(2xy-5x^2y)`

`= 6x^2y - 15x^3y`

`b, 2x^2y(xy-4xy^2+7y)`

`= 2x^3y^2 - 8x^3y^3 + 14x^2y^2`

`c, (-2/3xy^2 + 6yz^2).(-1/2xy)`

`= 1/3x^2y^3 - 3xy^2z^2`

Kaijo
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Lê Ngọc Bảo Ngân
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Nguyễn Lê Phước Thịnh
2 tháng 12 2023 lúc 20:09

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

\(=6x\left(x-y\right)+9xy-9y^2\)

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

\(=3\left(2x+3y\right)\left(x-y\right)\)

Bài 4:

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Nguyễn Ngọc Anh
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Tran Le Khanh Linh
19 tháng 4 2020 lúc 8:58

\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)

\(=\frac{2x}{x\left(x+y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2\left(x-2y\right)+x+2y+4}{\left(x+2y\right)\left(x-2y\right)}\)

\(=\frac{3x-2y+4}{\left(x+2y\right)\left(x-2y\right)}\)

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Nobi Nobita
19 tháng 4 2020 lúc 9:01

\(ĐKXĐ:\hept{\begin{cases}x\ne0\\y\ne0\\x\ne\pm2y\end{cases}}\)

\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}=\frac{2x}{x\left(x+2y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x+2y\right)\left(x-2y\right)}\)

\(=\frac{2}{x+2y}+\frac{1}{x-2y}+\frac{4}{\left(x+2y\right)\left(x-2y\right)}\)\(=\frac{2\left(x-2y\right)}{\left(x+2y\right)\left(x-2y\right)}+\frac{x+2y}{\left(x+2y\right)\left(x-2y\right)}+\frac{4}{\left(x+2y\right)\left(x-2y\right)}\)

\(=\frac{2\left(x-2y\right)+x+2y+4}{\left(x+2y\right)\left(x-2y\right)}=\frac{2x-4y+x+2y+4}{\left(x+2y\right)\left(x-2y\right)}\)

\(=\frac{3x-2y+4}{\left(x+2y\right)\left(x-2y\right)}\)

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Vũ Thị Hoa
19 tháng 4 2020 lúc 9:01

\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)

\(=\frac{2x}{x\left(x+2y\right)}+\frac{y}{y\left(x-2y\right)}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2x.y\left(x-2y\right)}{xy\left(x-2y\right)\left(x+2y\right)}+\frac{y.x\left(x+2y\right)}{xy\left(x-2y\right)\left(x+2y\right)}+\frac{4xy}{xy\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2x^2-4xy^2+x^2y+2xy^2+4xy}{xy\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{3x^2y-2xy^2+4xy}{xy\left(x-2y\right)\left(x+2y\right)}\)

P/s:#Học Tốt#

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Mtrangg
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Nguyễn Lê Phước Thịnh
4 tháng 8 2023 lúc 19:50

\(\dfrac{xy}{x-y}-\dfrac{2x^2}{y-2x}\)

\(=\dfrac{xy}{x-y}+\dfrac{2x^2}{2x-y}\)

\(=\dfrac{xy\left(2x-y\right)+2x^2\left(x-y\right)}{\left(x-y\right)\left(2x-y\right)}\)

\(=\dfrac{2x^2y-xy^2+2x^3-2x^2y}{\left(x-y\right)\left(2x-y\right)}\)

\(=\dfrac{2x^3-xy^2}{\left(x-y\right)\left(2x-y\right)}=\dfrac{x\left(2x^2-y^2\right)}{\left(x-y\right)\left(2x-y\right)}\)

Nguyễn Thanh Tùng
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Diệu Vy
20 tháng 12 2016 lúc 10:19

\(=\left[\frac{2xy}{\left(x-y\right).\left(x+y\right)}+\frac{x-y}{2.\left(x+y\right)}\right]:\frac{x+y}{2x}+\frac{x}{y-x}\)

\(=\frac{4xy+\left(x-y\right).\left(x-y\right)}{2.\left(x-y\right).\left(x+y\right)}.\frac{2x}{x+y}+\frac{x}{y-x}\)

\(=\frac{x^2+2xy+y^2}{\left(x-y\right).\left(x+y\right)^2}.x+\frac{x}{y-x}\)

\(=\frac{x.\left(x+y\right)^2}{\left(x-y\right).\left(x+y\right)^2}+\frac{x}{y-x}\)

\(=\frac{x}{x-y}-\frac{x}{x-y}=0\)

Bạn giùm mik nhé, tks bạn nhiều (:

Rocker Phong
12 tháng 8 2020 lúc 8:36

sai rồi

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