\(\left\{{}\begin{matrix}\left(x+y\right)^2-xy=19\\\left(x+y\right)-xy=-1\end{matrix}\right.\) \(\Rightarrow\left(x+y\right)^2-\left(x+y\right)-20=0\)

\(\Rightarrow\left[{}\begin{matrix}x+y=5\Rightarrow xy=6\\x+y=-4\Rightarrow xy=-3\end{matrix}\right.\)

- Với \(\left\{{}\begin{matrix}x+y=5\\xy=6\end{matrix}\right.\) theo Viet đảo x;y là nghiệm:

\(t^2-5t+6=0\Rightarrow\left[{}\begin{matrix}t=2\\t=3\end{matrix}\right.\)

\(\Rightarrow\left(x;y\right)=\left(2;3\right);\left(3;2\right)\)

- Với \(\left\{{}\begin{matrix}x+y=-4\\xy=-3\end{matrix}\right.\) theo Viet đảo x;y là nghiệm:

\(t^2+4t-3=0\Rightarrow\left[{}\begin{matrix}t=-2-\sqrt{7}\\t=-2+\sqrt{7}\end{matrix}\right.\)

\(\Rightarrow\left(x;y\right)=\left(-2-\sqrt{7};-2+\sqrt{7}\right);\left(-2+\sqrt{7};-2-\sqrt{7}\right)\)

\(x^3+y^3=\left(x^2+y^2\right)\sqrt{x^2-xy+y^2}\)

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

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

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

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

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

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

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

\(\Leftrightarrow\sqrt{4x-3}.\left(x-y\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{4x-3}=0\\\left(x-y\right)^2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}4x-3=0\\x-y=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{3}{4}\\x=y\end{cases}}\)

Xét trường hợp:

Với x=3/4

=>\(x=\frac{3}{4}\Leftrightarrow y.\frac{3}{4}=0\Leftrightarrow y=0\)

Với: \(x=y\)

Có: \(xy=\sqrt{4x-3}\Leftrightarrow x^2y^2=4x-3\Leftrightarrow x^4-4x+3=0\Leftrightarrow x\left(x^3-1\right)-3\left(x-1\right)=0\)\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x-1\right)+2x\left(x-1\right)+3\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-1\right)\left(x^2+2x+3\right)=0\)( vì x^2+2x+3 luôn dương. Tự c/m nhé )

\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)\(\Leftrightarrow x=y=1\)

KL:.................................

thanks anh ạ 

Nhân 3 lần vế pt 1 lên rồi trừ 2 vê cho nhau

x=7/9 thì y=7/2; x=0 thì y=0

4x + 3y = 5y - 5x

4x + 5x = 5y - 3y

9x        = 2y

Vậy x=2, y=9