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Đặt x+y=a; x-2y=b
=>6/a-3/b=3 và 1/a+7/b=2
=>a=5/3 và b=5
=>x+y=5/3 và x-2y=5
=>x=25/9; y=-10/9
a: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+1+1}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
=>x+1=1 và y-2=1/2
=>x=0 và y=5/2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x-2y}=\dfrac{1}{2}-\dfrac{1}{18}=\dfrac{9}{18}-\dfrac{1}{18}=\dfrac{8}{18}=\dfrac{4}{9}\\\dfrac{2}{2x-y}=\dfrac{1}{18}+\dfrac{1}{x-2y}\end{matrix}\right.\)
=>x-2y=9 và 2/2x-y=1/18+1/9=1/18+2/18=3/18=1/6
=>x-2y=9 và 2x-y=12
=>x=5; y=-2
c: \(\Leftrightarrow\left\{{}\begin{matrix}10\left|x-6\right|+15\left|y+1\right|=25\\10\left|x-6\right|-8\left|y+1\right|=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}23\left|y+1\right|=23\\\left|x-6\right|=1\end{matrix}\right.\)
=>|x-6|=1 và |y+1|=1
=>\(\left\{{}\begin{matrix}x\in\left\{7;5\right\}\\y\in\left\{0;-2\right\}\end{matrix}\right.\)
\(a.\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}-2=-1\\\dfrac{4}{x}+\dfrac{3}{y}-2=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a-b-2=-1\\4a+3b-2=5\end{matrix}\right.\) (với \(\dfrac{1}{x}=a-\dfrac{1}{y}=b\))
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{10}{7}\\b=\dfrac{3}{7}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{10}{7}\Rightarrow x=\dfrac{7}{10}\\\dfrac{1}{y}=\dfrac{3}{7}\Rightarrow y=\dfrac{7}{3}\end{matrix}\right.\)
\(b.\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{5}{\left(x+y\right)}=2\\\dfrac{3}{x}+\dfrac{1}{\left(x+y\right)}=\dfrac{17}{10}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2a+5b=2\\3a+b=\dfrac{17}{10}\end{matrix}\right.\) (với \(\dfrac{1}{x}=a-\dfrac{1}{x+y}=b\))
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=\dfrac{1}{5}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{2}\Rightarrow x=2\\\dfrac{1}{x+y}=\dfrac{1}{5}\Rightarrow y=3\end{matrix}\right.\)
\(c.\left\{{}\begin{matrix}\dfrac{2}{x-1}+\dfrac{1}{y+1}=7\\\dfrac{5}{x-1}-\dfrac{2}{y+1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a+b=7\\5a-2b=4\end{matrix}\right.\) (với \(\dfrac{1}{x-1}=a-\dfrac{1}{y+1}=b\))
\(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-1}=2\Rightarrow x=\dfrac{3}{2}\\\dfrac{1}{y+1}=3\Rightarrow y=-\dfrac{2}{3}\end{matrix}\right.\)
\(d.\left\{{}\begin{matrix}\dfrac{2}{\sqrt{x-1}}-\dfrac{1}{\sqrt{y-1}}=1\\\dfrac{1}{\sqrt{x-1}}+\dfrac{1}{\sqrt{y-1}}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a-b=1\\a+b=2\end{matrix}\right.\) (với \(\dfrac{1}{\sqrt{x-1}}=a-\dfrac{1}{\sqrt{y-1}}=b\))
\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x-1}}=1\Rightarrow x=2\\\dfrac{1}{\sqrt{y-1}}=1\Rightarrow y=2\end{matrix}\right.\)
\(\frac{1}{x-3}=a,\frac{1}{y-4}=b\)
\(hpt\Leftrightarrow\hept{\begin{cases}a+b=\frac{5}{3}\\4a-3b=\frac{3}{2}\end{cases}\Rightarrow\hept{\begin{cases}a=\frac{13}{14}\\b=\frac{31}{42}\end{cases}\Rightarrow}}\hept{\begin{cases}x=\frac{53}{13}\\y=\frac{166}{31}\end{cases}}\)
ĐKXĐ \(\hept{\begin{cases}x\ne2\\y\ne1\end{cases}}\)
Đặt \(\hept{\begin{cases}\frac{1}{x-2}=a\\\frac{1}{y-1}=b\end{cases}\left(a;b\ne0\right)}\)
Hệ trở thành \(\hept{\begin{cases}a+b=2\\2a-3b=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2a+2b=4\\2a-3b=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}5b=3\\a+b=2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}b=\frac{3}{5}\\a=\frac{7}{5}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x-2}=\frac{7}{5}\\\frac{1}{y-1}=\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-2=\frac{5}{7}\\y-1=\frac{5}{3}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{19}{7}\\y=\frac{8}{3}\end{cases}}\left(TmDKXD\right)\)
Đặt m = 1 / x - 3 và n = 1/y - 4
Khi đó ta có hệ m + n = 5/3
4 x x - 3 x n = 3/2
....Bạn tự giải tiếp nhé
Xem lại đề x + 3 hay x + 2