a/ Bạn tự giải

b/ ĐKXĐ:...

Cộng vế với vế: \(\frac{x-y}{y+12}=3\Rightarrow x-y=3y+36\Rightarrow x=4y+36\)

Thay vào pt đầu: \(\frac{4y+36}{y}-\frac{y}{y+12}=1\)
Đặt \(\frac{y+12}{y}=a\Rightarrow4a-\frac{1}{a}=1\Rightarrow4a^2-a-1=0\)

\(\Rightarrow a=\frac{1\pm\sqrt{17}}{8}\) \(\Rightarrow\frac{y+12}{y}=\frac{1\pm\sqrt{17}}{8}\)

\(\Rightarrow\left[{}\begin{matrix}y+12=y\left(\frac{1+\sqrt{17}}{8}\right)\\y+12=y\left(\frac{1-\sqrt{17}}{8}\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(\frac{-7+\sqrt{17}}{8}\right)y=12\\\left(\frac{-7-\sqrt{17}}{8}\right)y=12\end{matrix}\right.\) \(\Rightarrow y=...\)

Chắc bạn ghi sai đề, nghiệm quá xấu

3/ \(\Leftrightarrow\left\{{}\begin{matrix}3x^2+y^2=5\\3x^2-9y=3\end{matrix}\right.\) \(\Rightarrow y^2+9y=2\Rightarrow y^2+9y-2=0\Rightarrow y=...\)

4/ ĐKXĐ:...

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{3x-1}-3\sqrt{2y+1}=3\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)

\(\Rightarrow5\sqrt{3x-1}=15\Rightarrow\sqrt{3x-1}=3\Rightarrow x=\frac{10}{3}\)

\(\sqrt{2y+1}=\sqrt{3x-1}-1=3-1=2\Rightarrow2y+1=4\Rightarrow y=\frac{3}{2}\)

\(\left\{{}\begin{matrix}\frac{x}{y}-\frac{x}{x+12}=1\\-\frac{x}{y}+\frac{x}{x-12}=2\end{matrix}\right.\) \(\left(y\ne0;x\ne\pm12\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{24x}{x^2-144}=3\\y=\frac{x^2+12x}{12+2x}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=4+\sqrt{10}\\y=\frac{30+7\sqrt{10}}{10}\end{matrix}\right.\\\left\{{}\begin{matrix}x=4-\sqrt{10}\\y=\frac{30-7\sqrt{10}}{10}\end{matrix}\right.\end{matrix}\right.\)

Vậy....

Vì buồn ngủ quá nên ko bt có đúng không :3

\(e,\left\{{}\begin{matrix}\left(\frac{x}{y}\right)^3+\left(\frac{x}{y}\right)^2=12\\\left(xy\right)^2+xy=6\end{matrix}\right.\left(x;y\ne0\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy\in\left\{2;-3\right\}\end{matrix}\right.\)

Vì \(\frac{x}{y}=2>0\Rightarrow xy>0\Rightarrow xy=2\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2y\\2y^2=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\left(h\right)\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)

\(a,\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+\frac{x}{y}=3\\x+\frac{1}{y}+\frac{x}{y}=3\end{matrix}\right.\left(x;y\ne0\right)\)

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

Đặt \(\left\{{}\begin{matrix}x+\frac{1}{y}=a\\\frac{x}{y}=b\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a^2-b=3\\a+b=3\end{matrix}\right.\)

Làm nốt nha

\(\left\{{}\begin{matrix}\sqrt{x-1}+\sqrt{y-1}=2\\\frac{1}{x}+\frac{1}{y}=1\end{matrix}\right.\left(x;y\ge1\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y-2+2\sqrt{\left(x-1\right)\left(y-1\right)}=4\\x+y=xy\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+2\sqrt{xy-\left(x+y\right)+1}=6\\x+y=xy\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+2\sqrt{xy-xy+1}=6\\x+y=xy\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\xy=4\end{matrix}\right.\)

Làm nốt

Gửi bạn nhé

1

\(\left\{{}\begin{matrix}\frac{4}{x}+\frac{4}{y}=\frac{1}{3}\\\frac{4}{x}+\frac{6}{y}=\frac{2}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}=\frac{1}{12}\\\frac{2}{y}=\frac{2}{5}-\frac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{30}=\frac{1}{12}\\y=30\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{60}{7}\\y=30\end{matrix}\right.\)

Đặt: a=\(\frac{1}{x}\), b=\(\frac{1}{y}\) ta có:

\(\left\{{}\begin{matrix}a+b=\frac{1}{12}\\4a+6b=\frac{2}{5}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}4a+4b=\frac{1}{3}\\4a+6b=\frac{2}{5}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=\frac{1}{12}\\-2b=-\frac{1}{15}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{20}\\b=\frac{1}{30}\end{matrix}\right.\)

hay: \(\frac{1}{x}=\frac{1}{20}\Rightarrow x=20\)

\(\frac{1}{y}=\frac{1}{30}\Rightarrow y=30\)