Question

Giari hệ phương trình:

\(\left\{{}\begin{matrix}\dfrac{81}{x}+\dfrac{105}{y}=8\\\dfrac{54}{x}+\dfrac{42}{y}=4\end{matrix}\right.\)

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Answer 1

x,y khác 0 đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}81a+105b=8\\54a+42b=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}9.9.2a+105.2.b=8.2\\9.6.3a+42.3b=4.3\end{matrix}\right.\)

\(\Leftrightarrow\left(105.2-42.3\right)b=8.2-4.3=4\left(4-3\right)=4\)

\(\Leftrightarrow\left(105-21.3\right)b=2\)

\(\Leftrightarrow3\left(35-21\right)b=2\Rightarrow b=\dfrac{2}{3.14}=\dfrac{1}{3.7}=\dfrac{1}{21}\)

\(54a+42.\dfrac{1}{21}=4\Leftrightarrow54a+2=4\)

\(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\a=\dfrac{1}{27}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=27\\y=21\end{matrix}\right.\)