Đặt ẩn phụ nhé
\(\dfrac{1}{x+y}=a;\dfrac{1}{x-y}=b=< =>\int_{2a-3b=1}^{a+b=3}< =>\int_{2.\left(3-b\right)-3b=1}^{,a=3-b}< =>\int_{b=1}^{a=2}\)
<=>\(\dfrac{1}{x+y}=2;\dfrac{1}{x-y}=1< =>\int_{x-y=1}^{x+y=2}< =>\int_{y=0,5}^{x=1,5}\)
Đặt :
\(\left\{{}\begin{matrix}\dfrac{1}{x+y}=u\\\dfrac{1}{x-y}=v\end{matrix}\right.\)
Ta có hệ phương trình :
\(\left\{{}\begin{matrix}u+v=3\\2u-3v=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2u+2v=6\\2u-3v=1\end{matrix}\right.\)
\(\Leftrightarrow5v=5\Leftrightarrow v=1\)
Thay \(v=1\) vào phương trình thứ nhất ta đc :
\(u+1=3\Leftrightarrow u=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x+y}=2\\\dfrac{1}{x-y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=\dfrac{1}{2}\\x-y=1\end{matrix}\right.\)
\(\Leftrightarrow2y=-\dfrac{1}{2}\Rightarrow y=-\dfrac{1}{4}\)
Thay \(y=-\dfrac{1}{4}\) vào phương trình thứ 2 ta được :
\(x+\dfrac{1}{4}=1\Leftrightarrow x=\dfrac{3}{4}\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=-\dfrac{1}{4}\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}\dfrac{1}{x+y}+\dfrac{1}{x-y}=3\\\dfrac{2}{x+y}-\dfrac{3}{x-y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x-y+x+y}{\left(x+y\right)\left(x-y\right)}=3\\\dfrac{2x-2y+3x+3y}{\left(x+y\right)\left(x-y\right)}=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=3\left(x+y\right)\left(x-y\right)\\5x+y=\left(x+y\right)\left(x-y\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=3\left(x+y\right)\left(x-y\right)\\15x+3y=3\left(x+y\right)\left(x-y\right)\end{matrix}\right.\)
\(\Rightarrow2x=15x+3y\)
\(\Rightarrow15x+3y-2x=0\)
\(\Rightarrow13x+3y=0\)
\(\Rightarrow13x=-3y\Leftrightarrow x=-\dfrac{3}{13}y\)
Thay vào pt rồi tìm \(x;y\)
\(\left\{{}\begin{matrix}\dfrac{1}{x+y}+\dfrac{1}{x-y}=3\\\dfrac{2}{x+y}-\dfrac{3}{x-y}=1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{2x}{\left(x+y\right)\left(x-y\right)}=3\\\dfrac{2\left(x-y\right)-3\left(x+y\right)}{\left(x+y\right)\left(x-y\right)}=1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}2x=3\left(x+y\right)\left(x-y\right)\\-x-5y=\left(x+y\right)\left(x-y\right)\end{matrix}\right.\\ \Rightarrow\dfrac{2x}{3}=-x-5y\\ \Rightarrow\dfrac{2x}{3}+x=-5y\\ \Rightarrow\dfrac{5x}{3}=-5y\\ \Rightarrow\dfrac{x}{3}=-y\)
