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Λşαşşʝŋ GΩD
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hưng phúc
21 tháng 11 2021 lúc 22:01

1. \(\dfrac{x^3-4x^2+4x}{x^2-4}=\dfrac{x\left(x^2-4x+4\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}=\dfrac{x\left(x-2\right)}{x+2}\)

 

hưng phúc
21 tháng 11 2021 lúc 22:11

Đợi anh chút

Nguyễn Việt Lâm
21 tháng 11 2021 lúc 22:20

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

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

Thanh Sỹ
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Nguyễn Đức Trí
18 tháng 9 2023 lúc 23:41

a) \(\dfrac{x^3-1}{x^2+x+1}=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x^2+x+1}=x-1\)

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

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

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

c) \(\dfrac{ax^4-a^4x}{a^2+ax+x^2}\)

\(=\dfrac{ax\left(x^3-a^3\right)}{a^2+ax+x^2}\)

\(=\dfrac{ax\left(x-a\right)\left(a^2+ax+x^2\right)}{a^2+ax+x^2}\)

\(=ax\left(x-a\right)\)

....
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Nguyễn Hoàng Minh
17 tháng 12 2021 lúc 16:18

\(=\dfrac{x+2y}{\left(x-2y\right)\left(x+2y\right)}-\dfrac{x}{x+2}\cdot\dfrac{\left(x+2\right)^2}{x\left(x-2y\right)}=\dfrac{1}{x-2y}-\dfrac{x+2}{x-2y}=\dfrac{-x-1}{x-2y}\)

Lê Thị Thảo Uyên
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Nguyen Hong Thuy
21 tháng 10 2015 lúc 20:58

=(x+y)2-22/(x-y)(x+y)+4(x+y)

=(X+Y-2)(X+Y+2)/(X+Y)(X-Y+4)

Khánh Linh Đỗ
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HT.Phong (9A5)
26 tháng 10 2023 lúc 15:38

a) \(\dfrac{2\left(x+1\right)^2}{4x\left(x+1\right)}\left(x\ne0;x\ne-1\right)\)

\(=\dfrac{2\left(x+1\right)^2:2\left(x+1\right)}{4x\left(x+1\right):2\left(x+1\right)}\)

\(=\dfrac{x+1}{2x}\)

b) \(\dfrac{\left(8-x\right)\left(-x-2\right)}{\left(x+2\right)^2}\left(x\ne-2\right)\)

\(=\dfrac{-\left(8-x\right)\left(x+2\right)}{\left(x+2\right)^2}\)

\(=\dfrac{-\left(8-x\right)}{x+2}\)

\(=\dfrac{x-8}{x+2}\)

c) \(\dfrac{2\left(x-y\right)}{y-x}\left(x\ne y\right)\)

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

\(=-2\)

d) \(\dfrac{\left(x+2\right)^2}{2x+4}\left(x\ne-2\right)\)

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

\(=\dfrac{x+2}{2}\)

Toru
26 tháng 10 2023 lúc 15:38

ĐKXĐ: \(x\neq0;x\neq-1\)

\(\dfrac{2(x+1)^2}{4x(x+1)}=\dfrac{2(x+1)}{4x}=\dfrac{x+1}{2x}\)

$---$

ĐKXĐ: \(x\neq-2\)

\(\dfrac{(8-x)(-x-2)}{(x+2)^2}=\dfrac{-(8-x)(x+2)}{(x+2)^2}=\dfrac{x-8}{x+2}\)

$---$

ĐKXĐ: \(x\neq y\)

\(\dfrac{2(x-y)}{y-x}=\dfrac{-2(y-x)}{y-x}=-2\)

$---$

ĐKXĐ: \(x\neq-2\)

\(\dfrac{(x+2)^2}{2x+4}=\dfrac{(x+2)^2}{2(x+2)}=\dfrac{x+2}{2}\)

Lê Dương
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Nguyễn Thanh Hằng
16 tháng 11 2018 lúc 21:51

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

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

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

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

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

Vậy..

Quang Nguyễn
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Phan Nguyễn Duy Linh
10 tháng 9 2018 lúc 9:56

Có \(\frac{2}{x^2-y^2}\sqrt{\frac{9\left(x+2xy+y\right)}{4}}\) 

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

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

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

=\(\frac{2.3.\left(x+y\right)}{\left(x+y\right)\left(x-y\right).2}\) =\(\frac{3}{x-y}\)

ThanhNghiem
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HT.Phong (9A5)
7 tháng 10 2023 lúc 10:41

a) ĐKXĐ: \(x\ne2y,x\ne-y;x\ne-1\) 

b) \(B=\left(\dfrac{x-y}{2y-x}-\dfrac{x^2+y^2+y-2}{x^2-xy-2y^2}\right):\dfrac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\) 

\(B=\left[\dfrac{y-x}{x-2y}-\dfrac{x^2+y^2+y-2}{\left(x+y\right)\left(x-2y\right)}\right]:\dfrac{4x^4+4x^2y+y^2-4}{x\left(x+y\right)+\left(x+y\right)}\)

\(B=\left[\dfrac{\left(y-x\right)\left(x+y\right)}{\left(x-2y\right)\left(x+y\right)}-\dfrac{x^2+y^2+y-2}{\left(x+y\right)\left(x-2y\right)}\right]:\dfrac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+1\right)\left(x+y\right)}\)

\(B=\dfrac{y^2-x^2-x^2-y^2-y+2}{\left(x+y\right)\left(x-2y\right)}:\dfrac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+1\right)\left(x+y\right)}\)

\(B=\dfrac{-2x^2-y+2}{\left(x+y\right)\left(x-2y\right)}\cdot\dfrac{\left(x+1\right)\left(x+y\right)}{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}\)

\(B=\dfrac{-\left(2x^2+y-2\right)}{\left(x+y\right)\left(x-2y\right)}\cdot\dfrac{\left(x+1\right)\left(x+y\right)}{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}\)

\(B=\dfrac{-\left(x+1\right)}{\left(x-2y\right)\left(2x^2+y+2\right)}\)

trần công phúc
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Despacito
17 tháng 9 2017 lúc 14:40

b) \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

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

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

\(=\left(2x+y\right)\left(8x^2+2y^2\right)\)

\(=\left(2x+y\right)\left(4x+y\right).2xy\)