\(\left\{{}\begin{matrix}x+y=1500\\x+\dfrac{75}{100}x+y+\dfrac{68}{100}y=2583\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=1500\\\left(x+y\right)+\dfrac{3}{4}x+\dfrac{17}{25}y=2583\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=1500\\1500+\dfrac{3}{4}x+\dfrac{17}{25}y=2583\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=1500\\\dfrac{3}{4}x+\dfrac{17}{25}y=1083\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=900\\y=600\end{matrix}\right.\)

ĐKXĐ: \(x\notin\left\{0;-20\right\}\)

\(\left\{{}\begin{matrix}y=20+x\\\dfrac{100}{x}-\dfrac{100}{20+x}=\dfrac{5}{6}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=x+20\\\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=x+20\\\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=x+20\\\dfrac{400}{x\left(x+20\right)}=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=x+20\\x\left(x+20\right)=2400\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=x+20\\x^2+20x-2400=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=x+20\\\left(x+60\right)\left(x-40\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=x+20\\\left[{}\begin{matrix}x+60=0\\x-40=0\end{matrix}\right.\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=x+20\\x\in\left\{-60;40\right\}\end{matrix}\right.\)(nhận)

=>\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-60\\y=x+20=-60+20=-40\end{matrix}\right.\\\left\{{}\begin{matrix}x=40\\y=x+20=60\end{matrix}\right.\end{matrix}\right.\)

ĐKXĐ: ...

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+\dfrac{1}{x+y}+x-y+\dfrac{1}{x-y}=\dfrac{16}{3}\\\left(x+y\right)^2+\dfrac{1}{\left(x+y\right)^2}+\left(x-y\right)^2+\dfrac{1}{\left(x-y\right)^2}=\dfrac{100}{9}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+\dfrac{1}{x+y}+x-y+\dfrac{1}{x-y}=\dfrac{16}{3}\\\left(x+y+\dfrac{1}{x+y}\right)^2+\left(x-y+\dfrac{1}{x-y}\right)^2=\dfrac{136}{9}\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y+\dfrac{1}{x+y}=u\\x-y+\dfrac{1}{x-y}=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u+v=\dfrac{16}{3}\\u^2+v^2=\dfrac{136}{9}\end{matrix}\right.\)

Hệ cơ bản, chắc bạn tự giải quyết phần còn lại được

b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}+\dfrac{3}{y}=-3\\\dfrac{3}{x}-\dfrac{2}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y}=-10\\\dfrac{1}{x}+\dfrac{1}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)

Lời giải:

Phương hướng giải là bạn sử dụng phương pháp thế, biểu diễn $x$ theo $y$ qua 1 trong 2 PT, sau đó thế vô PT còn lại giải PT 1 ẩn $y$
a) \(\left\{\begin{matrix} x-6y=17\\ 5x+y=23\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=17+6y\\ 5x+y=23\end{matrix}\right.\)

\(\Rightarrow 5(17+6y)+y=23\)

\(\Leftrightarrow 31y=-62\Leftrightarrow y=-2\)

$x=17+6y=17+6(-2)=5$

Vậy $(x,y)=(5,-2)$

Các phần còn lại bạn giải tương tự

b) $(x,y)=(\frac{1}{4}, 0)$

c) $(x,y)=(3, 4)$

d) $(x,y)=(\frac{79}{21}, \frac{44}{21})$

a: \(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=5\\\dfrac{2}{x}-\dfrac{8}{y}=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y}=11\\\dfrac{1}{x}-\dfrac{4}{y}=-3\end{matrix}\right.\)

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

b: \(\left\{{}\begin{matrix}\dfrac{12}{x-3}-\dfrac{5}{y+2}=63\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{36}{x-3}-\dfrac{15}{y+2}=189\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{44}{x-3}=176\\\dfrac{8}{x-3}+\dfrac{15}{y+2}=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=\dfrac{1}{4}\\\dfrac{15}{y+2}=-13-\dfrac{8}{x-3}=-13-32=-45\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=-\dfrac{1}{3}-2=-\dfrac{7}{3}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{3}{y}=9\\\dfrac{7}{x}+\dfrac{4}{y}=17\end{matrix}\right.\)

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

\(\left\{{}\begin{matrix}4a+3b=9\\7a+4b=17\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}16a+12b=36\\21a+12b=51\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}-5a=-15\\4a+3b=9\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}a=3\\4.3+3b=9\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}a=3\\b=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=3\\\dfrac{1}{y}=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-1\end{matrix}\right.\)

Vậy ...

Ta có: \(\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{3}{y}=9\\\dfrac{7}{x}+\dfrac{4}{y}=17\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{28}{x}+\dfrac{21}{y}=63\\\dfrac{28}{x}+\dfrac{16}{y}=68\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y}=-5\\\dfrac{4}{x}+\dfrac{3}{y}=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{3}{-1}=9\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}-3=9\\y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}=12\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-1\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-1\end{matrix}\right.\)

Đặt \(a=\dfrac{1}{x};b=\dfrac{1}{y}\) ta được hệ : \(\left\{{}\begin{matrix}4a+3b=9\\7a+4b=17\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=-1\end{matrix}\right.\)

- Thay lại ta được : \(\left\{{}\begin{matrix}\dfrac{1}{x}=3\\\dfrac{1}{y}=-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-1\end{matrix}\right.\)

Vậy ...