\(\left\{{}\begin{matrix}x^2-xy-2y^2=0\\3x+y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x\left(1-3x\right)-2\left(1-3x\right)^2=0\\y=1-3x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-14x^2+11x-2=0\\y=1-3x\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{7}\end{matrix}\right.\\y=1-3x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{2}{7}\\y=\dfrac{1}{7}\end{matrix}\right.\end{matrix}\right.\)

Vậy...

`x^3+1=2y,y^3+1=2x`

`=>x^3-y^3=2y-2x`

`<=>(x-y)(x^2+xy+y^2)+2(x-y)=0`

`<=>(x-y)(x^2+xy+y^2+2)=0`

Vì `x^2+xy+y^2+2>=2>0`

`=>x-y=0<=>x=y` thay vào bthức

`=>x^3+1=2x`

`<=>x^3-2x+1=0`

`<=>x^3-x^2+x^2-2x+1=0`

`<=>x^2(x-1)+(x-1)^2=0`

`<=>(x-1)(x^2+x-1)=0`

`+)x=1=>x=y=1`

`+)x^2+x-1=0`

`\Delta=1+4=5`

`=>x_1=(-1-sqrt5)/2,x_2=(-1+sqrt5)/2`

`=>x=y=(-1-sqrt5)/2,x=y=z(-1+sqrt5)/2`

Vậy `(x,y)=(1,1),((-1-sqrt5)/2,(-1-sqrt5)/2),((-1+sqrt5)/2,(-1+sqrt5)/2)`

moi hok klop 6

\(\hept{\begin{cases}x+2y=3\\-2x-y=6\end{cases}< =>\hept{\begin{cases}x=3-2y\\-2\left(3-2y\right)-y\end{cases}< =>\hept{\begin{cases}x=3-2y\\-6+4y=6\end{cases}< =>\hept{\begin{cases}x=3-2y\\4y=12\end{cases}< =>\hept{\begin{cases}x=-3\\y=3\end{cases}}}}}}\)

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

Ta có: \(\hept{\begin{cases}\left(\frac{1}{x}+y\right)+\left(\frac{1}{x}-y\right)=\frac{5}{8}\\\left(\frac{1}{x}+y\right)-\left(\frac{1}{x}-y\right)=-\frac{3}{8}\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{2}{x}=\frac{5}{8}\\2y=-\frac{3}{8}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{16}{5}\\y=-\frac{3}{16}\end{cases}}}\)