Question

[๖²⁴ʱ乂ų✌й๏✌ρɾ๏༉]

giải hệ pt :

\(\dfrac{5}{x}+\dfrac{3}{y}=1\)

\(\dfrac{2}{x}+\dfrac{1}{y}=-1\)

Answer 1

[Nguyễn Lê Phước Thịnh]

ĐKXĐ: x<>0; y<>0

\(\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{3}{y}=1\\\dfrac{2}{x}+\dfrac{1}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x}+\dfrac{3}{y}=1\\\dfrac{6}{x}+\dfrac{3}{y}=-3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-\dfrac{1}{x}=4\\\dfrac{2}{x}+\dfrac{1}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\\dfrac{1}{y}=-1-\dfrac{2}{x}=-1-2:\dfrac{-1}{4}=-1+8=7\end{matrix}\right.\)

=>x=-1/4 và y=1/7

Answer 2

[datcoder]

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}a=\dfrac{1}{x}\\b=\dfrac{1}{y}\end{matrix}\right.\) 

Hệ phương trình trở thành \(\left\{{}\begin{matrix}5a+3b=1\\2a+b=-1\end{matrix}\right.\)

 \(\Rightarrow\left\{{}\begin{matrix}b=-1-2a\\5a+3\left(-1-2a\right)=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=-1-2a\\-a-3=1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a=-4\\b=-1-2.\left(-4\right)\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=-4\\b=7\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}a=\dfrac{1}{x}=-4\\b=\dfrac{1}{y}=7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\left(tm\right)\\y=\dfrac{1}{7}\left(tm\right)\end{matrix}\right.\)

Vậy HPT có nghiệm \(x=-\dfrac{1}{4}\) và \(y=\dfrac{1}{7}\)