Giải hệ phương trình: \(\left\{{}\begin{matrix}2\left(x+y\right)=5\left(x-y\right)\\\frac{20}{x+y}+\frac{20}{x-y}=7\end{matrix}\right.\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
DH
12 tháng 2 2020
\(\left\{{}\begin{matrix}2\left(x+y\right)=5\left(x-y\right)\\\frac{20}{x+y}+\frac{20}{x-y}=7\end{matrix}\right.\left(1\right)\) \(Đkxđ:x\ne\pm y\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{5}{x+y}=\frac{2}{x-y}\\\frac{20}{x+y}+\frac{20}{x-y}=7\end{matrix}\right.\left(2\right)\)
Đặt: \(\left\{{}\begin{matrix}a=\frac{1}{x+y}\\b=\frac{1}{x-y}\end{matrix}\right.\) Ta có hệ pt \((2)\) trở thành:
\(\left\{{}\begin{matrix}5a=2b\\20a+20b=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5a-2b=0\\20a+20b=7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}20a-8b=0\\20a+20b=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5a=2b\\28b=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1}{10}\\b=\frac{1}{4}\end{matrix}\right.\)
Với: \(\left\{{}\begin{matrix}a=\frac{1}{10}\\b=\frac{1}{4}\end{matrix}\right.\) Ta lại có hệ pt sau: \(\left\{{}\begin{matrix}\frac{1}{x+y}=\frac{1}{10}\\\frac{1}{x-y}=\frac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=10\\x-y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=14\\x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=3\end{matrix}\right.\left(tmđk\right)\)
Vậy ........
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 7 2019
1/ ĐKXĐ:...
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{x}+\frac{3}{y-2}=4\\\frac{12}{x}+\frac{3}{y-2}=3\end{matrix}\right.\) \(\Rightarrow\frac{10}{x}=-1\Rightarrow x=-10\)
\(\frac{4}{-10}+\frac{1}{y-2}=1\Rightarrow\frac{1}{y-2}=\frac{7}{5}\Rightarrow y-2=\frac{5}{7}\Rightarrow y=\frac{19}{7}\)
2/ ĐKXĐ:...
Đặt \(\left\{{}\begin{matrix}\frac{1}{2x-y}=a\\\frac{1}{x+y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a-b=0\\3a-6b=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{9}\\b=\frac{2}{9}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{2x-y}=\frac{1}{9}\\\frac{1}{x+y}=\frac{2}{9}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x-y=9\\x+y=\frac{9}{2}\end{matrix}\right.\) \(\Rightarrow...\)
3/ \(\Leftrightarrow\left\{{}\begin{matrix}5x+10y=3x-1\\2x+4=3x-6y-15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+10y=-1\\-x+6y=-19\end{matrix}\right.\) \(\Rightarrow...\)
4/ Bạn tự giải
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}x+y=a\\x-y=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2a=5b\\\frac{20}{a}+\frac{20}{b}=7\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=\frac{2a}{5}\\\frac{20}{a}+\frac{20}{b}=7\end{matrix}\right.\)
\(\Rightarrow\frac{20}{a}+\frac{50}{a}=7\)
\(\Rightarrow\frac{70}{a}=7\Rightarrow a=10\Rightarrow b=4\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=10\\x-y=4\end{matrix}\right.\) \(\left\{{}\begin{matrix}x=7\\y=3\end{matrix}\right.\)