\(\hept{\begin{cases}\frac{2}{x+y}+\frac{1}{x-y}=3\\\frac{1}{x+y}-\frac{3}{x-y}=1\end{cases}}\)

Đặt: \(u=\frac{1}{x+y};v=\frac{1}{x-y}\). Ta có: 

\(\hept{\begin{cases}2u+v=3\\u-3v=1\end{cases}}\)

\(\hept{\begin{cases}2u+v=3\\2u-6v=2\end{cases}}\)<=> 7v=1 => \(v=\frac{1}{7};u=\frac{10}{7}\)

\(< =>\hept{\begin{cases}\frac{1}{x+y}=\frac{10}{7}\\\frac{1}{x-y}=\frac{1}{7}\end{cases}}\) <=> \(\hept{\begin{cases}10x+10y=7\\x-y=7\end{cases}}\)<=> 10(y+7)+10y=7

<=> 20y+70=7

=> \(y=-\frac{63}{20}\); \(x=\frac{77}{20}\)

a = \(\frac{1}{x+y}\)

b = \(\frac{1}{x-y}\)

=>

\(\hept{\begin{cases}2a+b=3\\a-3b=1\end{cases}}\)

<=>

\(\hept{\begin{cases}2a+b=3\\2a-6b=2\end{cases}}\)

Trừ 2 vế PT

=> 7b = 1

=> b = 1/7

=> a = 10/7

=>

\(\hept{\begin{cases}x+y=\frac{7}{10}\\x-y=7\end{cases}}\)

<=>

\(\hept{\begin{cases}x=\frac{77}{20}\\y=-\frac{63}{20}\end{cases}}\)

<=> \(\hept{\begin{cases}\frac{2}{x}+\frac{4}{y}=\frac{2}{3}\\\frac{2}{x}-\frac{3}{y}=\frac{1}{4}\end{cases}}\)

<=> \(\hept{\begin{cases}\frac{7}{y}=\frac{5}{12}\\\frac{1}{x}+\frac{2}{y}=\frac{1}{3}\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{14}{3}\\y=\frac{84}{5}\end{cases}}\)

\(\begin{cases}\frac{x+y}{3}-\frac{x-y}{3}=\frac{14}{3}\left(1\right)\\3x-\frac{y}{2}+\frac{x}{4}=24\left(2\right)\end{cases}\).Từ \(\left(1\right)\Rightarrow\frac{2y}{3}=\frac{14}{3}\)

\(\Rightarrow2y=14\Rightarrow y=7\) thay vào (2) ta có:

\(3x-\frac{7}{2}+\frac{x}{4}=24\Rightarrow3x+\frac{x}{4}=24+\frac{7}{2}\)

\(\Rightarrow\frac{13x}{4}=\frac{55}{2}\Rightarrow13x\cdot2=55\cdot4\)

\(\Rightarrow26x=220\Rightarrow x=\frac{220}{26}=\frac{110}{13}\)

Vậy hệ pt có nghiệm là \(x=\frac{110}{13};y=7\)

\(\int^{1+\frac{4}{x-1}-\frac{3}{y-1}=2}_{\frac{4}{x-1}+\frac{2}{y-1}=6}\Leftrightarrow\int^{\frac{1}{y-1}=1}_{\frac{1}{x-1}=1}\Leftrightarrow x=y=2\)

( đặt ẩn phụ cho dẽ làm nhá)

Bạn viết sai dấu y+1 = y-1 ( theo bạn nhẩm 2;2)

Đặt \(x+1=u;y-2=v\)

Hệ trở thành \(\hept{\begin{cases}\frac{2}{u}+\frac{1}{v}=\frac{1}{3}\\\frac{3}{u}+\frac{2}{v}=\frac{1}{5}\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{4}{u}+\frac{2}{v}=\frac{2}{3}\left(1\right)\\\frac{3}{u}+\frac{2}{v}=\frac{1}{5}\left(2\right)\end{cases}}\)

Lấy (1) - (2), ta được\(\frac{1}{u}=\frac{7}{15}\Leftrightarrow u=\frac{15}{7}\)

\(\Rightarrow x=\frac{15}{7}-1=\frac{8}{7}\)

Từ đó tính được \(y=\frac{1}{3}\)

Vậy hệ có 1 nghiệm \(\left(\frac{8}{7};\frac{1}{3}\right)\)

<=> \(\hept{\begin{cases}\frac{4}{x+1}+\frac{2}{y-2}=\frac{2}{3}\\\frac{3}{x+1}+\frac{2}{y-2}=\frac{1}{5}\end{cases}}\)

<=> \(\hept{\begin{cases}\frac{1}{x+1}=\frac{7}{15}\\\frac{3}{x+1}+\frac{2}{y-2}=\frac{1}{5}\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{8}{7}\\y=\frac{7}{5}\end{cases}}\)

Bài này đơn giản thôi :))

\(\text{HPT}\Leftrightarrow \left\{\begin{matrix} \frac{x+y}{xy}=\frac{3}{2}\\ \frac{y+z}{yz}=\frac{2}{3}\\ \frac{x+z}{xz}=\frac{7}{6}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{1}{x}+\frac{1}{y}=\frac{3}{2}\\ \frac{1}{y}+\frac{1}{z}=\frac{2}{3}\\ \frac{1}{x}+\frac{1}{z}=\frac{7}{6}\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} \frac{2}{x}=\frac{3}{2}+\frac{7}{6}-\frac{2}{3}\\ \frac{2}{y}=\frac{3}{2}+\frac{2}{3}-\frac{7}{6}\\ \frac{2}{z}=\frac{2}{3}+\frac{7}{6}-\frac{3}{2}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1\\ y=2\\ z=6\end{matrix}\right.\)

Vậy $(x,y,z)=(1,2,6)$ là nghiệm của hệ phương trình

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{3}+\frac{y}{3}=\frac{7}{3}\\\frac{11}{12}x-\frac{y}{6}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{6}+\frac{y}{6}=\frac{7}{6}\left(1\right)\\\frac{11x}{12}-\frac{y}{6}=1\left(2\right)\end{matrix}\right.\)

Lấy (1) cộng (2) ta được

\(\frac{13}{12}x=\frac{13}{6}\Rightarrow x=2\Rightarrow y=5\)

Vậy.............

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=7\\8x-2y+3x=12\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}13x=26\\y=7-x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x+2y=14\\11x-2y=12\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\)

Vậy ....