1.(\(\dfrac{x^2+xy}{x^3+x^2y+xy^2+y^3}+\dfrac{y}{x^2+y^2}\)) :(\(\dfrac{1}{x-y}-\dfrac{2xy}{x^3-x^2y+xy^2-y^3}\))
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
KN
Kien Nguyen
18 tháng 12 2017
Đúng(0)
Những câu hỏi liên quan
QN
1
28 tháng 11 2022
\(=\dfrac{x\left(x+y\right)}{\left(x+y\right)\left(x^2+y^2\right)}\cdot\left(\dfrac{1}{x-y}-\dfrac{2xy}{\left(x-y\right)\left(x^2+y^2\right)}\right)\)
\(=\dfrac{x}{x^2+y^2}\cdot\dfrac{x^2+y^2-2xy}{\left(x-y\right)\left(x^2+y^2\right)}\)
\(=\dfrac{x}{x^2+y^2}\cdot\dfrac{x-y}{x^2+y^2}=\dfrac{x\left(x-y\right)}{\left(x^2+y^2\right)^2}\)
12 tháng 12 2022
\(=\left(\dfrac{x\left(x+y\right)}{x^2\left(x+y\right)+y^2\left(x+y\right)}+\dfrac{y}{x^2+y^2}\right):\left(\dfrac{1}{x-y}-\dfrac{2xy}{x^2\left(x-y\right)+y^2\left(x-y\right)}\right)\)
\(=\dfrac{x+y}{x^2+y^2}:\left(\dfrac{1}{x-y}-\dfrac{2xy}{\left(x-y\right)\left(x^2+y^2\right)}\right)\)
\(=\dfrac{x+y}{x^2+y^2}:\dfrac{x^2+y^2-2xy}{\left(x-y\right)\left(x^2+y^2\right)}\)
\(=\dfrac{x+y}{x^2+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+y^2\right)}{\left(x-y\right)^2}\)
\(=\dfrac{x+y}{x-y}\)
7 tháng 12 2017
a,\(\frac{x^2+y^2-xy}{x^2-y^2}:\frac{x^3+y^3}{x^2+y^2-2xy} =\frac{x^2+y^2-xy}{(x-y)(x+y)}\frac{(x+y)^2}{(x+y) (x^2-xy+y^2)}=\frac{1}{x-y} \)
b,\(\frac{x^3y+xy^3}{x^4y}:(x^2+y^2)=\frac{xy(x^2+y^2)}{x^4y(x^2+y^2)}=\frac{1}{x^3} \)
c,\(\frac{x^2-xy}{y}:\frac{x^2-xy}{xy+y}:\frac{x^2-1}{x^2+y} =\frac{x(x-y)y(x+y)(x^2+y)}{yx(x-y)(x^2-1)} =\frac{(x^2+y)(x+y)}{x^2-1} \)
d,\(\frac{x^2+y}{y}:(\frac{z}{x^2}:\frac{xy}{x^2y})=\frac{x^2+y}{ y}:(\frac{z}{x^2}\frac{x^2y}{xy})=\frac{x^2+y}{y}\frac{z}{x} \)
13 tháng 6 2022
a: \(=\dfrac{4x^2+4x+1-\left(4x^2-4x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{2x+1}\cdot\dfrac{5}{4x}=\dfrac{10}{2x+1}\)
c: \(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
QD
20 tháng 8 2017
a)\(\dfrac{2x^2-10xy}{2xy}+\dfrac{5y-x}{y}+\dfrac{x+2y}{x}\)
\(=\dfrac{2x\left(x-5y\right)}{2xy}+\dfrac{5y-x}{y}+\dfrac{x+2y}{x}\)
\(=\dfrac{x-5y}{y}+\dfrac{5y-x}{y}+\dfrac{x+2y}{x}\)
\(=\dfrac{x\left(x-5y\right)+x\left(5y-x\right)+y\left(x+2y\right)}{xy}\)
\(=\dfrac{x^2-5xy+5xy-x^2+xy+2y^2}{xy}\)
\(=\dfrac{y\left(x+2y\right)}{xy}\)
NN
24 tháng 11 2017
b) \(\dfrac{x+1}{2x-2}+\dfrac{x^2+3}{2-2x^2}\)
\(=\dfrac{x+1}{2x-2}-\dfrac{x^2+3}{2x^2-2}\)
\(=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{x^2+3}{2\left(x^2-1\right)}\)
\(=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{x^2+3}{2\left(x-1\right)\left(x+1\right)}\) MTC: \(2\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{x^2+3}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)-\left(x^2+3\right)}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2-x^2-3}{2\left(x-1\right)\left(x+1\right)}\)
e) \(\dfrac{2x^2-xy}{x-y}+\dfrac{xy+y^2}{y-x}+\dfrac{2y^2-x^2}{x-y}\)
\(=\dfrac{2x^2-xy}{x-y}-\dfrac{xy+y^2}{x-y}+\dfrac{2y^2-x^2}{x-y}\)
\(=\dfrac{\left(2x^2-xy\right)-\left(xy+y^2\right)+\left(2y^2-x^2\right)}{x-y}\)
\(=\dfrac{2x^2-xy-xy-y^2+2y^2-x^2}{x-y}\)
\(=\dfrac{x^2-2xy+y^2}{x-y}\)
\(=\dfrac{\left(x-y\right)^2}{x-y}\)
\(=x-y\)
PT
11 tháng 12 2017
a) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{\left(2x+1\right)\left(2x+1\right)}{2x^2-1}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{2x^2-1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{2x^2-1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{\left(2x+1-2x-1\right)\left(2x+1+2x-1\right)}{2x^2-1}\right):\dfrac{4x}{10x-5}\)
\(=\dfrac{4x}{2x^2-1}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{5}{2x+1}\)
b) \(\left(\dfrac{1}{x^2+1}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1}{x^2+1}-\dfrac{x\left(2-x\right)}{x\left(x+1\right)}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{1}{x}+\dfrac{x^2}{x}-\dfrac{2x}{x}\right)\)
\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{x^2-2x+1}{x}\right)\)
\(=\dfrac{\left(x-1\right)^2}{x^2+1}.\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x}{x^2+1}\)
c) d) Tự làm đi mình làm biếng quass >.< ^^