\(\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{6}{2x}+\dfrac{6}{y}=\dfrac{1}{4}\)
\(\Leftrightarrow6\left(\dfrac{1}{2x}+\dfrac{1}{y}\right)=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2x}+\dfrac{1}{y}=\dfrac{1}{24}^{\left(1\right)}\)
Lại có: \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}^{\left(2\right)}\)
Lấy (2) trừ (1) ta có:
\(\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{1}{2x}-\dfrac{1}{y}=\dfrac{1}{16}-\dfrac{1}{24}\)
\(\Leftrightarrow\dfrac{2-1}{2x}=\dfrac{1}{48}\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{48}\)
=> 2x = 48
<=> x = 24
Thay x = 24 vào (2) ta có:
\(\dfrac{1}{24}+\dfrac{1}{y}=\dfrac{1}{16}\)
\(\Leftrightarrow\dfrac{1}{y}=\dfrac{1}{48}\)
=> y = 48
Vậy ...
Ta có: \(\dfrac{3}{x}\) + \(\dfrac{6}{y}\) = \(\dfrac{1}{4}\)
<=> 3(\(\dfrac{1}{x}\) + \(\dfrac{2}{y}\) ) = \(\dfrac{1}{4}\)
<=> \(\dfrac{1}{x}\) + \(\dfrac{2}{y}\) = \(\dfrac{1}{12}\) (1)
Mặt khác: \(\dfrac{1}{x}\) + \(\dfrac{1}{y}\) = \(\dfrac{1}{16}\) (2)
Trừ (2) cho (1) vế theo vế ta được:
\(\dfrac{1}{x}\) + \(\dfrac{2}{y}\) - \(\dfrac{1}{x}\) - \(\dfrac{1}{y}\) = \(\dfrac{1}{12}\) - \(\dfrac{1}{16}\)
<=> \(\dfrac{1}{y}\) = \(\dfrac{1}{48}\) <=> y = 48
Thay y =48 vào (2) ta có: \(\dfrac{1}{x}\) + \(\dfrac{1}{48}\) = \(\dfrac{1}{16}\)
<=> \(\dfrac{1}{x}\) = \(\dfrac{1}{24}\) <=> x = 24
Vậy x =24 ; y =48
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}\\\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\end{matrix}\right.\)\((I)\)
Đặt \(a=\dfrac{1}{x};b=\dfrac{1}{y}\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=\dfrac{1}{16}\\3a+6b=\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=\dfrac{1}{16}\\a+2b=\dfrac{1}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=\dfrac{1}{16}\\b=\dfrac{1}{48}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{24}\\b=\dfrac{1}{48}\end{matrix}\right.\)\((II)\)
Thay \(a=\dfrac{1}{x};b=\dfrac{1}{y}\) vào \((II)\), ta được:\(\left\{{}\begin{matrix}x=24\\y=48\end{matrix}\right.\)
Vậy hệ phương trính (I) có nghiệm \(\left(x;y\right)=\left(24;48\right)\)
