Question

Giải phương trình 

\(\hept{\begin{cases}\left(x-y+1\right)^2=1\\2x\left(x+y\right)=1\\x#y,y#0\end{cases}}\)

Answer 1

\(\left(1\right)\Rightarrow\orbr{\begin{cases}x=y\\x=y-2\end{cases}}\)

\(\left(2\right)\Rightarrow4y^3=1\Rightarrow x=y=\sqrt[3]{\frac{1}{4}}\) Hoặc \(4y\left(y-1\right)=1\Rightarrow\left(y-\frac{1}{2}\right)^2=\frac{1}{2}\Rightarrow\orbr{\begin{cases}y=\frac{1-\sqrt{2}}{2}\\y=\frac{1+\sqrt{2}}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-3-\sqrt{2}}{2}\\\frac{-3+\sqrt{2}}{2}\end{cases}}\)