Đặt: \(\left\{{}\begin{matrix}\dfrac{x}{x+1}=a\\\dfrac{y}{y-3}=b\end{matrix}\right.\) Hệ: \(\left\{{}\begin{matrix}5a+b=27\\2a-3b=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=5\\b=2\end{matrix}\right.\)
Thay a và b ngược lên tính ra được x và y
C2: Thay vì đặt trực tiếp \(\dfrac{x}{x+1}\) và \(\dfrac{y}{y-3}\) rất khó tính toán thì mk sẽ rút gọn r mới đặt
\(\left\{{}\begin{matrix}\dfrac{5x}{x+1}+\dfrac{y}{y-3}=27\\\dfrac{2x}{x+1}-\dfrac{3y}{y-3}=4\end{matrix}\right.\) (x \(\ne\) -1; y \(\ne\) 3)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}\dfrac{5x+5-5}{x+1}+\dfrac{y-3+3}{y-3}=27\\\dfrac{2x+2-2}{x+1}-\dfrac{3y-9+9}{y-3}=4\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}5-\dfrac{5}{x+1}+1+\dfrac{3}{y-3}=27\\2-\dfrac{2}{x+1}-3-\dfrac{9}{y-3}=4\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}\dfrac{-5}{x+1}+\dfrac{3}{y-3}=21\\\dfrac{-2}{x+1}-\dfrac{9}{y-3}=5\end{matrix}\right.\)
Đặt \(\dfrac{1}{x+1}=a\); \(\dfrac{1}{y-3}=b\)
Khi đó hpt trở thành:
\(\left\{{}\begin{matrix}-5a+3b=21\\-2a-9b=5\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-15a+9b=63\\-2a-9b=5\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-17a=68\\-2a-9b=5\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=-4\\-2\cdot\left(-4\right)-9b=5\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=-4\\9b=3\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=-4\\b=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}\dfrac{1}{x+1}=-4\\\dfrac{1}{y-3}=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-4\left(x+1\right)=1\\y-3=3\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}-4x-4=1\\y=6\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{-5}{4}\\y=6\end{matrix}\right.\) (TM)
Vậy hpt trên có nghiệm duy nhất (x; y) = (\(\dfrac{-5}{4}\); 6)
Chúc bn học tốt!
Ta có: \(\left\{{}\begin{matrix}\dfrac{5x}{x+1}+\dfrac{y}{y-3}=27\\\dfrac{2x}{x+1}-\dfrac{3y}{y-3}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5x+5}{x+1}-\dfrac{5}{x+1}+\dfrac{y-3}{y-3}+\dfrac{3}{y-3}=27\\\dfrac{2x+2}{x+1}-\dfrac{2}{x+1}-\dfrac{3y-9}{y-3}-\dfrac{9}{y-3}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5+1-\dfrac{5}{x+1}+\dfrac{3}{y-3}=27\\2-3-\dfrac{2}{x+1}-\dfrac{9}{y-3}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-5}{x+1}+\dfrac{3}{y-3}=21\\\dfrac{-2}{x+1}-\dfrac{9}{y-3}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-10}{x+1}+\dfrac{6}{y-3}=42\\\dfrac{-10}{x+1}-\dfrac{45}{y-3}=25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{51}{y-3}=17\\\dfrac{-2}{x+1}-\dfrac{9}{y-3}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-3=3\\\dfrac{-2}{x+1}-\dfrac{9}{3}=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=6\\\dfrac{-2}{x+1}=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1=\dfrac{-1}{4}\\y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{4}\\y=6\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{5}{4}\\y=6\end{matrix}\right.\)
