\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)

\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x+y\right)}\)

\(=\frac{30x\left(x-y\right)-5x\left(x+y\right)}{5\left(x+y\right).10\left(x+y\right)}\)

\(=\frac{5x\left(5x-7y\right)}{50\left(x+y\right)\left(x-y\right)}\)

\(=\frac{x\left(5x-7y\right)}{\left(x+y\right)\left(x-y\right)}\)

chỗ cuối tớ sai 

\(=\frac{x\left(5x-7y\right)}{10\left(x+y\right)\left(x-y\right)}\)

đây nha , e xin lỗi

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

                                                          \(=\frac{3\left(x-1\right)+\left(2x-1\right)-2.2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)\left(x+1\right)}\)

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

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

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

b) \(\frac{3x}{5x+5y}-\frac{x}{10x-10y}=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x-y\right)}\)

                                                   \(=\frac{3x.10\left(x-y\right)-x.5\left(x+y\right)}{50\left(x-y\right)\left(x+y\right)}\)

                                                   \(=\frac{30x\left(x-y\right)+5x\left(x+y\right)}{50\left(x-y\right)\left(x+y\right)}\)

                                                   \(=\frac{5x\left[6\left(x-y\right)-\left(x+y\right)\right]}{50\left(x-y\right)\left(x+y\right)}\)

                                                   \(=\frac{5x\left(5x-7y\right)}{50\left(x-y\right)\left(x+y\right)}\)

                                                   \(=\frac{x\left(5x-7y\right)}{10\left(x-y\right)\left(x+y\right)}\)

c) \(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}=\frac{5x^2-y-x\left(3x-2y\right)}{xy}\)

                                                \(=\frac{5x^2-y-3x^2+2xy}{xy}\)

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

d) \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)

                                            \(=\frac{3x-x+6}{2x\left(x+3\right)}\)

                                            \(=\frac{2x+6}{2x\left(x+3\right)}\)

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

                                            \(=\frac{2}{2x}=\frac{1}{x}\) 

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

P/s:#Học Tốt#

b) (ko chép lại đề nhé)  \(=\frac{x^2\left(x-y\right)^2}{\left(x+y\right)\left(x-y\right)}\cdot\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{xy\left(x^2-xy+y^2\right)}=\frac{x\left(x-y\right)}{y}\)

Đơn thức đầu tiên trong mẫu của phân thức thứ 2 có lẽ là \(x^3y\) 

xin loi em khong biet!

a) .....\(=\frac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}\cdot\frac{2\left(x^2+x+1\right)}{x+1}=\frac{2\left(x^2+x+1\right)}{x-1}\)(đến đây ko biết đã tối giản chưa nữa)

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

b: \(\dfrac{3x^2-x}{x-1}+\dfrac{x+2}{1-x}+\dfrac{3-2x^2}{x-1}\)

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

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

\(\frac{y}{2x^2-xy}+\frac{4x}{y^2-2xy}=0\)

<=>\(\frac{y}{x\left(2x-y\right)}-\frac{4x}{y\left(2x-y\right)}=0\)

<=>\(\frac{y^2}{xy\left(2x-y\right)}-\frac{4x^2}{xy\left(2x-y\right)}=0\)

 =>y²-(2x)²=0

<=>(y-2x)(y+2x)=0

<=>y-2x=0 hoặc y+2x=0

M chỉ làm đc đến đó thôi!!!!!

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

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

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

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

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

Bạn giùm mik nhé, tks bạn nhiều (:

sai rồi