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

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

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

\(PT\left(1\right)\Leftrightarrow x=m+1-my\\ PT\left(2\right)\Leftrightarrow m^2+m-m^2y+y=3m-1\\ \Leftrightarrow y\left(1-m^2\right)=-m^2+2m-1\\ \Leftrightarrow y=\dfrac{\left(m-1\right)^2}{\left(m-1\right)\left(m+1\right)}=\dfrac{m-1}{m+1}\\ \Leftrightarrow x=m+1-\dfrac{m\left(m-1\right)}{m+1}=\dfrac{m^2+2m+1-m^2+m}{m+1}=\dfrac{3m+1}{m+1}\)

Vậy \(\left(x;y\right)=\left(\dfrac{3m+1}{m+1};\dfrac{m-1}{m+1}\right)\)

\(\left\{{}\begin{matrix}2xy=30\\x^2-y^2=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{15}{y}\\\left(\frac{15}{y}\right)^2-y^2=16\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{15}{y}\\\frac{225}{y^2}-y^2=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{15}{y}\\225-y^4=16y^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{15}{y}\\y^4+16y^2-225=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{15}{y}\\\left(y^2-9\right)\left(y^2+25\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{15}{y}\\\left[{}\begin{matrix}y^2=9\\y^2=-25\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{15}{3}=5\\y=3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}2x+3y=-2\\3x-2y=-3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-9y=-6\\6x-4y=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5y=0\\2x+3y=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=0\\2x+3.0=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\2x=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=-\frac{3}{2}\end{matrix}\right.\)

Vậy hệ phương trình có nghiệm duy nhất là \(\left(x;y\right)=\left(-\frac{3}{2};0\right)\)

\(\Leftrightarrow\hept{\begin{cases}x^2+3xy=10\\x^2+xy+4y^2+3xy=16\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(x+2y\right)^2=16\\x^2+3xy=10\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+2y=4\Rightarrow x=4-2y\\x^2+3xy=10\end{cases}}\) hoặc \(\hept{\begin{cases}x+2y=-4\Rightarrow x=-4-2y\\x^2+3xy=10\end{cases}}\)

Xong thế x=4-2y hoặc -4-2y vào phương trình x^2 +3xy=10 thành phương trình bậc 2 một ẩn, GPT=> x,y

\(\Leftrightarrow\hept{\begin{cases}\orbr{\begin{cases}\\\end{cases}}\\\end{cases}}\)

Nhận xét: 

+) Nếu x = 0 => y = 0 ; z = 0 => (0;0;0) là nghiệm của hệ

+) Nếu x khác 0 => y ;z khác 0.khi đó

hệ <=> 

3/2 = (x+y)/xy

5/6 = (z+y)/yz

7/8 = (x+z)/xz

<=> 3/2 = 1/y + 1/x  (1)

5/6 = 1/y + 1/z  (2)

7/8 = 1/z + 1/x  (3)

Cộng tưng vế của 3 pt trên ta được 2.(1/x + 1/y + 1/z) = 77/24 => 1/x + 1/y + 1/z = 77/48

Từ (1) => 1/z = 77/48 - 3/2 = ... => z = ...

Tương tự => x ; y

3xy=2(x+y)

2xy-xy=2x+2y

2xy-2x=2y+xy

2x(y-1)=(2+x)y

(y-1)/y=(2+x)/2x(1)

5yz=6(z+y)

4yz+yz=4z+4y+2z+2y

4yz-4z=4y-yz+2z+2y

................................................

còn lại thì giao hoán, rồi chịu

\(\begin{cases}x^3=7x+3y\left(1\right)\\y^3=7y+3x\left(2\right)\end{cases}\). Lấy \(\left(1\right)-\left(2\right)\) ta được

\(\left(x-y\right)\left(x^2+xy+y^2-4\right)=0\)

​\(x^3=7x+3x\Leftrightarrow x^3=10x\)

\(\Leftrightarrow x^3-10x=0\Leftrightarrow x\left(x^2-10\right)=0\)\(\Leftrightarrow\begin{cases}x=y=0\\x=y=\pm\sqrt{10}\end{cases}\)

\(\begin{cases}x^2+xy+y^2=4\\x^3+y^3=10\left(x+y\right)\end{cases}\) đặt \(\begin{cases}S=x+y\\P=xy\end{cases}\) \(\left(S^2\ge4P\right)\) ta có:

\(\begin{cases}P=S^2-4\\S^3-3SP-10S=0\end{cases}\) thay \(P=S^2-4\) ta có:

\(S^3-3S\left(S^2-4\right)-10S=0\)

\(\Leftrightarrow-2S\left(S-1\right)\left(S+1\right)=0\)\(\Leftrightarrow\left[\begin{array}{nghiempt}S=0\\S=1\\S=-1\end{array}\right.\)