\(\left\{{}\begin{matrix}x+y=1000\\\dfrac{15}{100}x+\dfrac{17}{100}y=1162\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{15}{100}x+\dfrac{15}{100}y=150\\\dfrac{15}{100}x+\dfrac{17}{100}y=1162\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{2}{100}y=150-1162=-1012\\x+y=1000\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=50600\\x=1000-50600=-49600\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+y=1500\\x+\dfrac{75}{100}x+y+\dfrac{68}{100}y=2583\end{matrix}\right.\)
\(\left\{{}\begin{matrix}y=20+x\\\dfrac{100}{x}-\dfrac{100}{20+x}=\dfrac{5}{6}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+2y=10\\\dfrac{1}{2}x-y=1\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{1}{y}=2\\\dfrac{6}{x}-\dfrac{2}{y}=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=10\\\dfrac{1}{x}-\dfrac{2}{y}=15\end{matrix}\right.\)
Giải hpt
a)\(\left\{{}\begin{matrix}\dfrac{2x}{x+1}+\dfrac{y}{y+1}=2\\\dfrac{x}{x+1}+\dfrac{3}{y+1}=-1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{x}{y}-\dfrac{x}{y+12}=1\\\dfrac{x}{y+12}-\dfrac{x}{y}=2\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\dfrac{x+y}{xy}+\dfrac{xy}{x+y}=\dfrac{5}{2}\\\dfrac{x-y}{xy}+\dfrac{xy}{x-y}=\dfrac{10}{3}\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{2x}{y-1}+\dfrac{3y}{x-1}=1\\\dfrac{2y}{x-1}-\dfrac{5x}{y-1}=2\end{matrix}\right.\)
Giải hệ phương trình:
a) \(\left\{{}\begin{matrix}\dfrac{x}{35}-y=2\\y-\dfrac{x}{50}=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}\\\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\end{matrix}\right.\)
(1) giải hpt:
a) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}^{ }\\\dfrac{8}{x}+\dfrac{5}{y}=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{x-1}{2}-y=1\\2x+y=1\end{matrix}\right.\)
giúp mk vs ak
(1) giải hpt:
a) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{8}{x}+\dfrac{5}{y}=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\dfrac{x-1}{2}-y=1\\2x+y=1\end{matrix}\right.\)
giúp mk vs ạ
Giải hệ phương trình sau:
a. \(\left\{{}\begin{matrix}\dfrac{x+2}{y}=\dfrac{x+1}{y-2}\\\dfrac{5x+1}{5x-2}=\dfrac{y-2}{y+2}\end{matrix}\right.\)
b. \(\left\{{}\begin{matrix}2x+\left|y\right|=4\\4x-3y=1\end{matrix}\right.\)
