a, \(\left\{{}\begin{matrix}35x-28y=21\\35x-45y=40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}17y=-19\\x=\dfrac{3+4y}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{19}{17}\\x=-\dfrac{13}{17}\end{matrix}\right.\)
b, Đặt x;y khác 0
Đặt 1/x = t ; 1/y = u
\(\left\{{}\begin{matrix}t-8u=18\\5t+4u=51\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5t-40u=90\\5t+4u=51\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-44u=39\\t=18+8u\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u=-\dfrac{39}{44}\\t=\dfrac{120}{11}\end{matrix}\right.\)
Theo cách đặt y = -44/39 ; x = 11/120 (tm)
\(a,\\ \Leftrightarrow\left\{{}\begin{matrix}35x-28y=21_{\left(1\right)}\\35x-45y=40_{\left(2\right)}\end{matrix}\right.\\ Lấy\left(1\right)-\left(2\right),ta.đc:\\ -17y=19\Leftrightarrow y=\dfrac{-19}{17}\\ Thay.vào.\left(1\right):\\ 35x-28.\dfrac{-19}{17}=21\Leftrightarrow x=\dfrac{-5}{17}\)
Vậy ......
\(b,\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=18+\dfrac{1}{y}\\5.\left(18+\dfrac{1}{y}\right)+\dfrac{4}{y}=51\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=18+\dfrac{1}{y}\\\dfrac{9}{y}=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=18-\dfrac{13}{3}\\\dfrac{1}{y}=\dfrac{-13}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{41}{3}\\\dfrac{1}{y}=\dfrac{-13}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{41}\\y=\dfrac{-3}{13}\end{matrix}\right.\)
câu này với ạ! Mình cảm ơn.
