ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\frac{1}{x}=u\\\frac{1}{y}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\frac{3}{5}u+v=10\\\frac{3}{4}u+\frac{3}{4}v=12\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3u+5v=50\\3u+3v=48\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}v=1\\u=15\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}=15\\\frac{1}{y}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{15}\\y=1\end{matrix}\right.\)

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne3\\y\ne1\end{matrix}\right.\)

Đặt `(x)/(x-3)` = a, `(y)/(y-1)` = b

  \(\text{Hệ}\Leftrightarrow\left\{{}\begin{matrix}a+3b=5\\4a-b=7\end{matrix}\right.\\ \Leftrightarrow...\\ \Leftrightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x-3}=2\\\dfrac{y}{y-1}=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=2x-6\\y=y-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=6\\-1=0\left(vô.lí\right)\end{matrix}\right.\)

Vậy hpt vô nghiệm

Đặt $x^2 = a > 0$ và $y^2 = b > 0$ thì hệ đã cho trở thành:

$\left\{\begin{aligned}&4a - 3b = 5\\&a + 2b = 4\\ \end{aligned}\right. \Leftrightarrow \left\{\begin{aligned}&4a - 3b = 5\\&a = 4 - 2b\\ \end{aligned}\right. \Leftrightarrow \left\{\begin{aligned}&16 - 8b - 3b = 5\\&a = 4 - 2b\\ \end{aligned}\right.$

$ \Leftrightarrow \left\{\begin{aligned}&- 11b = -11\\&a = 4 - 2b\\ \end{aligned}\right. \Leftrightarrow \left\{\begin{aligned}&b = 1 \, (tm)\\&a = 2 \, (tm)\\ \end{aligned}\right.$

Suy ra $x^2 = 2$ và $y^2 = 1$ từ đó em suy ra các nghiệm $(x;y)$ nhé

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

Lời giải:

Đặt $(x-3)^2=a$. Khi đó pt đã cho tương đương với:

$(x^2-6x+9-9)^2+13(x-3)^2-77=0$

$\Leftrightarrow [(x-3)^2-9]^2+13(x-3)^2-77=0$

$\Leftrightarrow (a-9)^2+13a-77=0$

$\Leftrightarrow a^2-5a+4=0$

$\Leftrightarrow (a-1)(a-4)=0$

$\Leftrightarroe a=1$ hoặc $a=4$

Đến đây thì đơn giản rồi.

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x-1+2}{x-1}+\frac{3\left(y+2\right)-6}{y+2}=7\\\frac{2}{x-1}-\frac{5}{y+2}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}1+\frac{2}{x-1}+3-\frac{6}{y+2}=7\\\frac{2}{x-1}-\frac{5}{y+2}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{x-1}-\frac{6}{y+2}=3\\\frac{2}{x-1}-\frac{5}{y+2}=4\end{matrix}\right.\)

đặt \(\left\{{}\begin{matrix}a=\frac{1}{x-1}\\b=\frac{1}{y+2}\end{matrix}\right.\) ta có : \(\left\{{}\begin{matrix}2a-6b=3\\2a-5b=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2a=6b+3\\b=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\frac{9}{2}\\b=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x-1}=\frac{9}{2}\\\frac{1}{y+2}=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=\frac{2}{9}\\y+2=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{11}{9}\\y=-1\end{matrix}\right.\)