Question

[Nguyễn Duy Hưng]

giải hệ phương trình bằng phương pháp đặt ẩn phụ

Answer 1

[HT.Phong (9A5)]

\(\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{3}{2}+\dfrac{1}{y}\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{24}\end{matrix}\right.\) (Đk: x,y ≠ 0) 

Đặt: \(\dfrac{1}{x}=u;\dfrac{1}{y}=v\) 

Hệ trở thành:

\(\left\{{}\begin{matrix}u=\dfrac{3}{2}+v\\u+v=\dfrac{1}{24}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u=\dfrac{3}{2}+v\\\dfrac{3}{2}+v+v=\dfrac{1}{24}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u=\dfrac{3}{2}+v\\2v=-\dfrac{35}{24}\end{matrix}\right.\) 

\(\Leftrightarrow\left\{{}\begin{matrix}u=\dfrac{37}{48}\\v=-\dfrac{35}{48}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{37}{48}\\\dfrac{1}{y}=\dfrac{-35}{48}\end{matrix}\right.\) 

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{48}{37}\\y=-\dfrac{48}{35}\end{matrix}\right.\)

Vậy: \(\left(x;y\right)=\left(\dfrac{48}{37};-\dfrac{48}{35}\right)\)