Question

giải hệ \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{80}\) và \(\dfrac{10}{x}+\dfrac{12}{y}=\dfrac{2}{15}\)

Answer 1

Đặt \(\dfrac{1}{x}\)= a; \(\dfrac{1}{y}\)= b, ta có hệ phương trình:
\(\left\{{}\begin{matrix}a+b=\dfrac{1}{80}\\10a+12b=\dfrac{2}{15}\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}10a+10b=\dfrac{1}{8}\\10a+12b=\dfrac{2}{15}\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}-2b=-\dfrac{1}{120}\\a+b=\dfrac{1}{80}\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}b=\dfrac{1}{240}\\a=\dfrac{1}{80}-\dfrac{1}{240}=\dfrac{1}{120}\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{120}\\\dfrac{1}{y}=\dfrac{1}{240}\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=120\\y=240\end{matrix}\right.\)