Đặt : \(\left\{{}\begin{matrix}\frac{1}{x}=a\\\frac{1}{y}=b\end{matrix}\right.\)ta có hệ :

\(\left\{{}\begin{matrix}a+b=\frac{1}{6}\\3a+4b=\frac{17}{30}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b=\frac{1}{2}\\3a+4b=\frac{17}{30}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=\frac{17}{30}-\frac{1}{2}=\frac{1}{15}\\a=\frac{\frac{1}{2}-\frac{1}{5}}{3}=\frac{1}{10}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=15\\y=10\end{matrix}\right.\)

Vậy ...

\(1,\dfrac{x-1}{3}=x+1\\ \Leftrightarrow x-1=3x+3\\ \Leftrightarrow3x-x=3+1\\ \Leftrightarrow x=2\)

PT có tập nghiệm S = {2}

\(2,\sqrt{16x^2+8x+1}-2=x\\ \Leftrightarrow\sqrt{\left(4x+1\right)^2}-2=x\\\Leftrightarrow 4x+1-2=x\\ \Leftrightarrow4x-x=2-1\\ \Leftrightarrow x=\dfrac{1}{3}\)

PT có tập nghiệm S = {1/3}

\(3,\left\{{}\begin{matrix}2x+y=17\\x-2y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+y=17\\2x-4y=2\end{matrix}\right.\\ \Leftrightarrow\left(2x+y\right)-\left(2x-4y\right)=17-2\\ \Leftrightarrow5y=15\\ \Leftrightarrow y=3\\ \Leftrightarrow2x+3=17\\ \Leftrightarrow2x=14\\ \Leftrightarrow x=7\)

PTHH có tập nghiệm (x; y) là (7; 3)

\(a.\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}-2=-1\\\dfrac{4}{x}+\dfrac{3}{y}-2=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a-b-2=-1\\4a+3b-2=5\end{matrix}\right.\) (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 \(\dfrac{1}{\sqrt{x-1}}=a-\dfrac{1}{\sqrt{y-1}}=b\))

\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x-1}}=1\Rightarrow x=2\\\dfrac{1}{\sqrt{y-1}}=1\Rightarrow y=2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{x+y}{5}=\dfrac{x-y}{3}\\\dfrac{x}{4}=\dfrac{y}{2}+1\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}3x+3y=5x-5y\\x=2y+4\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}2x-8y=0\\x-2y=4\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x-4y=0\\x-2y=4\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}y=2\\x=8\end{matrix}\right.\)

Trả lời:

theo đề bài: x^2+y^2 = -1

-> phương trình vô nghiệm do x^2+y^2 >=0 nên không thể tìm được x,y thỏa điều kiện đề bài.