Đáp án: `(x;y)=(\frac{-1}{4};\frac{1}{7})`
Giải thích các bước giải:
$ĐKXĐ:x;y\neq0$
Đặt `\frac{1}{x}=a;\frac{1}{y}=b`
Khi đó, ta có hệ: $\large \left \{ {{5a+3b=1} \atop {2a+b=-1}} \right.$
$⇔\large \left \{ {{5a+3b=1} \atop {b=-1-2a}} \right.⇔\large \left \{ {{5a+3(-1-2a)=1} \atop {b=-1-2a}} \right.$
$⇔\large \left \{ {{5a-3-6a=1} \atop {b=-1-2a}} \right.⇔\large \left \{ {{-a-3=1} \atop {b=-1-2a}} \right.⇔\large \left \{ {{-a=4} \atop {b=-1-2a}} \right.$
$⇔\large \left \{ {{a=-4} \atop {b=7}} \right.⇔\large \left \{ {{\frac{1}{x}=-4} \atop {\frac{1}{y}=7}} \right.$
$⇔\large \left \{ {{x=\frac{-1}{4}} \atop {y=\frac{1}{7}}} \right.$
