Ta có: \(\left\{{}\begin{matrix}2x-3y=1\\-4x+6y=2\end{matrix}\right.\)

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

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

Vậy: Hệ phương trình vô nghiệm

\(\begin{cases} x-3y=-2\\2x+y=3 \end{cases} <=> \begin{cases} x-3(3-2x)=-2\\y=3-2x \end{cases} <=> \begin{cases} 7x=7\\y=3-2x \end{cases} \\<=> \begin{cases} x=1\\y=3-2.1 \end{cases} <=>\begin{cases} x=1\\y=1 \end{cases}\)

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

\(\hept{\begin{cases}x-y=m\left(1\right)\Rightarrow y=x-m\\2x+y=4\left(2\right)\end{cases}}\)

Thay vào (2) => 2x+(x-m)=4

\(\Leftrightarrow\hept{\begin{cases}y=x-m\\3x-m-4=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=x-m\\x=\frac{4+m}{3}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{4+m}{3}\\y=\frac{4-m}{3}-m=\frac{4-4m}{3}\end{cases}}}\)

\(\hept{\begin{cases}x-y=m\\2x+y=4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x-y+2x+y=m+4\\2x+y=4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}3x=m+4\\2x+y=4\end{cases}}\)                                                                                                                                                                               \(\Leftrightarrow\)   \(\hept{\begin{cases}x=\frac{m+4}{3}\\2.\frac{m+4}{3}+y=4\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{m+4}{3}\\\frac{2m+8}{3}+y=4\end{cases}}\)                                                                                                                                                                       \(\Leftrightarrow\hept{\begin{cases}x=\frac{m+4}{3}\\y=\frac{4-2m}{3}\end{cases}}\)                                      Vậy hệ pt có nghiệm duy nhất là: \(\left(x;y\right)=\left(\frac{m+4}{3};\frac{4-2m}{3}\right)\)    

1.

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y+x^3y+xy^2+xy=-\dfrac{5}{4}\\x^4+y^2+xy\left(1+2x\right)=-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y\right)+xy+xy\left(x^2+y\right)=-\dfrac{5}{4}\\\left(x^2+y\right)^2+xy=-\dfrac{5}{4}\end{matrix}\right.\left(1\right)\)

Đặt \(\left\{{}\begin{matrix}x^2+y=a\\xy=b\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a+b+ab=-\dfrac{5}{4}\\a^2+b=-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-a^2-\dfrac{5}{4}-a\left(a^2+\dfrac{5}{4}\right)=-\dfrac{5}{4}\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2-a^3-\dfrac{1}{4}a=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-a\left(a^2-a+\dfrac{1}{4}\right)=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a\left(a-\dfrac{1}{2}\right)^2=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=0\\xy=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\sqrt[3]{10}}{2}\\y=-\dfrac{5}{2\sqrt[3]{10}}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=\dfrac{1}{2}\\xy=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-\dfrac{3}{2}\end{matrix}\right.\)

Kết luận: Phương trình đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(\dfrac{\sqrt[3]{10}}{2};-\dfrac{5}{2\sqrt[3]{10}}\right);\left(1;-\dfrac{3}{2}\right)\right\}\)

2.

\(\left\{{}\begin{matrix}\left(x+1\right)^3-16\left(x+1\right)=\left(\dfrac{2}{y}\right)^3-4\left(\dfrac{2}{y}\right)\\1+\left(\dfrac{2}{y}\right)^2=5\left(x+1\right)^2+5\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+1=u\\\dfrac{2}{y}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^3-16u=v^3-4v\\v^2=5u^2+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u^3-v^3=16u-4v\\4=v^2-5u^2\end{matrix}\right.\)

\(\Rightarrow4\left(u^3-v^3\right)=\left(16u-4v\right)\left(v^2-5u^2\right)\)

\(\Leftrightarrow21u^3-5u^2v-4uv^2=0\)

\(\Leftrightarrow u\left(7u-4v\right)\left(3u+v\right)=0\Rightarrow\left[{}\begin{matrix}u=0\Rightarrow v^2=4\\u=\dfrac{4v}{7}\Rightarrow4=v^2-5\left(\dfrac{4v}{7}\right)^2\\v=-3u\Rightarrow4=\left(-3u\right)^2-5u^2\end{matrix}\right.\) 

\(\Rightarrow...\)

9: \(\left\{{}\begin{matrix}3x-2=y\\2x+3y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-y=2\\2x+3y=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6x-2y=4\\6x+9y=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11y=-14\\3x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{14}{11}\\x=\dfrac{y+2}{3}=\dfrac{\dfrac{14}{11}+2}{3}=\dfrac{12}{11}\end{matrix}\right.\)

\(9,\Leftrightarrow\left\{{}\begin{matrix}3x-2=y\\2x+3\left(3x-2\right)=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-2=y\\11x=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{12}{11}\\y=\dfrac{14}{11}\end{matrix}\right.\)

\(10,\Leftrightarrow\left\{{}\begin{matrix}2x=2-3y\\2\left(2-3y\right)-y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=2-3y\\4-6y-y-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{14}\\y=\dfrac{3}{7}\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x+3-x-3y-3=0\\x=3+y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3y=0\\x=3+y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=3+y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=3+0=3\end{matrix}\right.\)