Đặt \(\dfrac{1}{x+y-1}=a;\dfrac{1}{2x-y+3}=b\)
Hệ phương trình trở thành:
\(\left\{{}\begin{matrix}4a-5b=\dfrac{5}{3}\\3a+b=\dfrac{7}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12a-15b=5\\12a+4b=\dfrac{28}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-19b=\dfrac{-3}{5}\\3a+b=\dfrac{7}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{3}{95}\\a=\dfrac{26}{57}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+y-1}=\dfrac{26}{57}\\\dfrac{1}{2x-y+3}=\dfrac{3}{95}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y-1=\dfrac{57}{26}\\2x-y+3=\dfrac{95}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=\dfrac{83}{26}\\2x-y=\dfrac{86}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x=\dfrac{2485}{78}\\x+y=\dfrac{83}{26}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2485}{234}\\y=\dfrac{83}{26}-\dfrac{2485}{234}=\dfrac{-869}{117}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left(x,y\right)=\left(\dfrac{2485}{234};\dfrac{-869}{117}\right)\)