Question

\(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}=\frac{1}{72}\\\frac{1}{x}+\frac{1}{z}=\frac{1}{63}\\\frac{1}{y}+\frac{1}{z}=\frac{1}{56}\end{matrix}\right.\)

Answer 1

Đặt \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\)

\(hpt\Leftrightarrow\left\{{}\begin{matrix}a+b=\frac{1}{72}\\a+c=\frac{1}{63}\\b+c=\frac{1}{56}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=\frac{1}{72}\\a-b=\frac{1}{63}-\frac{1}{56}\\b+c=\frac{1}{56}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1}{72}-b\\2b=\frac{1}{72}+\frac{1}{504}\\c=\frac{1}{56}-b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1}{72}-b\\b=\frac{\frac{1}{72}+\frac{1}{504}}{2}=\frac{1}{126}\\c=\frac{1}{56}-b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1}{72}-\frac{1}{126}=\frac{1}{168}\\b=\frac{1}{126}\\c=\frac{1}{56}-\frac{1}{126}=\frac{5}{504}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=168\\y=126\\z=100,8\end{matrix}\right.\)