\(\hept{\begin{cases}x^2-2y^2=-1\left(1\right)\\2x^3-y^3=2y-x\end{cases}}\)

\(\Rightarrow\left(2x^3-y^2\right)\cdot1=\left(x^2-2y^2\right)\left(2y-x\right)\)(nhân chéo 2 vế để cùng bậc)

\(\Rightarrow2x^3-y^3=2x^2y-x^3-4y^3+2xy^2\)

\(\Rightarrow3x^3-2x^2y-2xy^2+3y^3=0\)

\(\Rightarrow3\left(x+y\right)\left(x^2-xy+y^2\right)-2xy\left(x+y\right)=0\)

\(\Rightarrow\left(x+y\right)\left(3x^2-5xy+3y^2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+y=0\\x=y=0\end{cases}\Rightarrow x=-y}\)

Thay x=-y vào (1): \(x^2-2x^2=-1\Rightarrow x^2=1\Rightarrow\orbr{\begin{cases}x=1\Rightarrow y=-1\\x=-1\Rightarrow y=1\end{cases}}\)

\(HPT\Leftrightarrow\hept{\begin{cases}\left(x+y\right)^2-2xy=11\\\left(x+y\right)+xy=3+4\sqrt{2}\end{cases}}\)

Đặt x+y=a;xy=b thì hệ trở thành: 

\(\hept{\begin{cases}a^2-2b=11\\a+b=3+4\sqrt{2}\end{cases}\Leftrightarrow\hept{\begin{cases}a^2-2b=11\\b=3+4\sqrt{2}-a\end{cases}}}\)

=> \(a^2-2\left(3+4\sqrt{2}-a\right)=11\)

<=>\(a^2-6-8\sqrt{2}+2a-11=0\)

\(\Leftrightarrow a^2+2a-17-8\sqrt{2}=0\)

\(\Leftrightarrow a^2+2a-\left(16+2.4.\sqrt{2}+2-1\right)=0\)

\(\Leftrightarrow\left(a+1\right)^2-\left(4+\sqrt{2}\right)^2=0\)

\(\Leftrightarrow\left(a+1-4-\sqrt{2}\right)\left(a+1+4+\sqrt{2}\right)=0\)

\(\Leftrightarrow\left(a-3-\sqrt{2}\right)\left(a+5+\sqrt{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a-3-\sqrt{2}=0\\a+5+\sqrt{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=3+\sqrt{2}\\a=-5-\sqrt{2}\end{cases}}}\)

=>\(\orbr{\begin{cases}b=3+4\sqrt{2}-3-\sqrt{2}=3\sqrt{2}\\b=3+4\sqrt{2}+5+\sqrt{2}=8+5\sqrt{2}\end{cases}}\)

- Với \(a=3+\sqrt{2},b=3\sqrt{2}\),ta có: \(x+y=3+\sqrt{2}\Rightarrow y=3+\sqrt{2}-x\) (1)

Thay (1) vào \(xy=3\sqrt{2}=b\Rightarrow x\left(3+\sqrt{2}-x\right)=3\sqrt{2}\)

\(\Leftrightarrow3x+x\sqrt{2}-x^2=3\sqrt{2}\Leftrightarrow x\left(3-x\right)+\sqrt{2}\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\sqrt{2}-x\right)=0\Leftrightarrow\orbr{\begin{cases}x=3\\x=\sqrt{2}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\sqrt{2}\\y=3\end{cases}}\)

- Với \(a=-5-\sqrt{2},b=8+5\sqrt{2}\), ta có: \(x+y=-5-\sqrt{2}\Rightarrow y=-5-\sqrt{2}-x\)(2)

Thay (2) vào \(xy=8+5\sqrt{2}=b\Rightarrow x\left(-5-\sqrt{2}-x\right)=8+5\sqrt{2}\)

\(\Leftrightarrow-x^2-5x-x\sqrt{2}-8-5\sqrt{2}=0\)

\(\Leftrightarrow-x^2+\left(-5-\sqrt{2}\right)x+\left(-8-5\sqrt{2}\right)=0\)(3)

\(\Delta=\left(-5-\sqrt{2}\right)^2-4.\left(-1\right).\left(-8-5\sqrt{2}\right)\)

\(=27+10\sqrt{2}-32-20\sqrt{2}=-5-10\sqrt{2}< 0\)

=>pt (3) vô nghiệm

Vậy \(\left(x;y\right)=\left(3;\sqrt{2}\right)\) hoaojwc \(\left(\sqrt{2};3\right)\)

1) \(\hept{\begin{cases}x^2+y^2-xy=1\\x+x^2y=2y^3\end{cases}\Leftrightarrow}\hept{\begin{cases}x^2+y^2=1+xy\\x\left(1+xy\right)=2y^3\end{cases}\Rightarrow x\left(x^2+y^2\right)=2y^3}\)

\(\Leftrightarrow\left(x^3-y^3\right)+\left(xy^2-y^3\right)=0\Leftrightarrow\left(x-y\right)\left(x^2+y^2+xy\right)+y^2\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+2y^2+xy\right)=0\Leftrightarrow\orbr{\begin{cases}x=y\\x^2+2y^2+xy=0\end{cases}}\)

+) \(x=y\Rightarrow\hept{\begin{cases}y^2+y^2-y^2=1\\y+y^3=2y^3\end{cases}\Rightarrow}x=y=\pm1\)

+) \(x^2+2y^2+xy=0\)Vì y=0 không là nghiệm của hệ nên ta chia 2 vế phương trình cho y²:

\(\Rightarrow\left(\frac{x}{y}\right)^2+\frac{x}{y}+2=0\)( Vô nghiệm)

Vậy hệ có nghiệm (1;1),(-1;-1).

2/ \(\hept{\begin{cases}x+y=\sqrt{x+3y}\\x^2+y^2+xy=3\end{cases}\Rightarrow\hept{\begin{cases}x^2+y^2+2xy=x+3y\\x^2+y^2+xy=3\end{cases}}}\Rightarrow xy=x+3y-3\)

\(\Leftrightarrow\left(x-xy\right)+\left(3y-3\right)\Leftrightarrow\left(x-3\right)\left(1-y\right)=0\Leftrightarrow\orbr{\begin{cases}x=3\Rightarrow y\in\varnothing\\y=1\Rightarrow x=1\end{cases}}\)

Vậy hệ có nghiệm (1;1).

1) 

\(\hept{\begin{cases}\left(\sqrt{2}+\sqrt{3}\right)x-y\sqrt{2}=\sqrt{2}\\\left(\sqrt{2}+\sqrt{3}\right)x+y\sqrt{3}=-\sqrt{3}\end{cases}\Leftrightarrow\hept{\begin{cases}-y\left(\sqrt{2}+\sqrt{3}\right)=\sqrt{2}+\sqrt{3}\\\left(\sqrt{2}+\sqrt{3}\right)x+y\sqrt{3}=-\sqrt{3}\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\y=-1\end{cases}}\)

Bó tay. com

ko bít sorry nhaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa