Question

Gỉai hệ phương trình sau đây :

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

Answer 1

Lời giải:

\(\left\{\begin{matrix} \frac{1}{x}+\frac{1}{y}=\frac{1}{24}(1)\\ \frac{1}{x}-\frac{3}{2y}=0(2)\end{matrix}\right.\)

Lấy PT(1) trừ PT(2) ta có:

\(\left(\frac{1}{x}+\frac{1}{y}\right)-\left(\frac{1}{x}-\frac{3}{2y}\right)=\frac{1}{24}\)

\(\Leftrightarrow \frac{1}{y}(1+\frac{3}{2})=\frac{1}{24}\)

\(\Leftrightarrow \frac{1}{y}.\frac{5}{2}=\frac{1}{24}\Rightarrow y=60\)

Thay vào PT(1): \(\frac{1}{x}=\frac{1}{24}-\frac{1}{y}=\frac{1}{24}-\frac{1}{60}=\frac{1}{40}\)

\(\Rightarrow x=40\)

Vậy \((x,y)=(40,60)\)