K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

giải hệ phương trình 1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\) 2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\) 3 ,...
Đọc tiếp

giải hệ phương trình

1 , \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y-1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)

2, \(\left\{{}\begin{matrix}2\left(\frac{1}{x}+\frac{1}{2y}\right)+3\left(\frac{1}{x}-\frac{1}{2y}\right)^2=9\\\left(\frac{1}{x}+\frac{1}{2y}\right)-6\left(\frac{1}{x}-\frac{1}{2y}\right)^2=-3\end{matrix}\right.\)

3 , \(\left\{{}\begin{matrix}\frac{xy}{x+y}=\frac{2}{3}\\\frac{yz}{y+z}=\frac{6}{5}\\\frac{zx}{z+x}=\frac{3}{4}\end{matrix}\right.\)

4 , \(\left\{{}\begin{matrix}2xy-3\frac{x}{y}=15\\xy+\frac{x}{y}=15\end{matrix}\right.\)

5 , \(\left\{{}\begin{matrix}x+y+3xy=5\\x^2+y^2=1\end{matrix}\right.\)

6 , \(\left\{{}\begin{matrix}x+y+xy=11\\x^2+y^2+3\left(x+y\right)=28\end{matrix}\right.\)

7, \(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=4\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\end{matrix}\right.\)

8, \(\left\{{}\begin{matrix}x+y+xy=11\\xy\left(x+y\right)=30\end{matrix}\right.\)

9 , \(\left\{{}\begin{matrix}x^5+y^5=1\\x^9+y^9=x^4+y^4\end{matrix}\right.\)

3
26 tháng 8 2017

KON 'NICHIWA ON" NANOKO: chào cô

22 tháng 2 2017

ĐK: !x! khác !y!

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

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

\(B=\frac{x^3+x^2y-xy^2-y^3}{MSC}=\frac{x^2\left(x+y\right)-y^2\left(x+y\right)}{MSC}=\frac{\left(x+y\right)^2\left(x-y\right)}{\left(x-y\right)^2\left(x+y\right)^2}=\frac{1}{x-y}\)