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

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

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

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

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

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

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

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

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

ĐKXĐ: \(xy\ne0;x\ne\pm y\)

\(\left\{{}\begin{matrix}\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{\dfrac{1}{y}+\dfrac{1}{x}}=\dfrac{5}{2}\\\dfrac{1}{y}-\dfrac{1}{x}+\dfrac{1}{\dfrac{1}{y}-\dfrac{1}{x}}=\dfrac{10}{3}\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}a+b+\dfrac{1}{a+b}=\dfrac{5}{2}\\b-a+\dfrac{1}{b-a}=\dfrac{10}{3}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)^2-\dfrac{5}{2}\left(a+b\right)+1=0\\\left(b-a\right)^2-\dfrac{10}{3}\left(b-a\right)+1=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}a+b=2\\a+b=\dfrac{1}{2}\end{matrix}\right.\\\left[{}\begin{matrix}b-a=3\\b-a=\dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}a+b=2\\b-a=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=\dfrac{5}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=\dfrac{5}{2}\end{matrix}\right.\)

3 TH còn lại xét tương tự

Bạn cần làm gì với biểu thức này?
 

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

=>Hai phân thức này bằng nhau

b: \(\dfrac{xy+y}{x}=\dfrac{y\left(x+1\right)}{x}\)

\(\dfrac{xy+x}{y}=\dfrac{x\left(y+1\right)}{y}\)

Vì \(\dfrac{y\left(x+1\right)}{x}\ne\dfrac{x\left(y+1\right)}{y}\)

nên hai phân thức này không bằng nhau

c: \(\dfrac{-6}{4y}=\dfrac{-6:2}{4y:2}=\dfrac{-3}{2y}\)

\(\dfrac{3y}{-2y^2}=\dfrac{-3y}{2y^2}=\dfrac{-3y}{y\cdot2y}=\dfrac{-3}{2y}\)

Do đó: \(\dfrac{-6}{4y}=\dfrac{3y}{-2y^2}\)

=>Hai phân thức này bằng nhau

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

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

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

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

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

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,\(\frac{x^2+y}{y}:(\frac{z}{x^2}:\frac{xy}{x^2y})=\frac{x^2+y}{ y}:(\frac{z}{x^2}\frac{x^2y}{xy})=\frac{x^2+y}{y}\frac{z}{x} \)

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

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

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

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

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

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

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

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

mình viết trên máy tinh hơi xấu bạn thông cảm nhé!!!Nếu ko chê có thể xem cách giải này!