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

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

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

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

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

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

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

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

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

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

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 

b, \(x^3+3x^2y-4y^3+x-y=0\)

\(\Leftrightarrow x^3-x^2y+4x^2y-4xy^2+4xy^2-4y^3+x-y=0\)

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

\(\Leftrightarrow\left(x-y\right)\left(x^2+4xy+4y^2+1\right)=0\)

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

Khi đó pt (2) của hệ trở thành: 

\(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)

\(\Leftrightarrow\left(x^2+5x+5\right)^2-1=24\)

\(\Leftrightarrow\left(x^2+5x+5\right)^2-5^2=0\)

\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+10\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

Vậy hệ có nghiệm \(\left(x;y\right)\in\left\{\left(0;0\right),\left(-5;-5\right)\right\}\)

t ms lên 7  =>  I don't know  

\(\hept{\begin{cases}7\left(2x+y\right)-5\left(3x+y\right)=6\\3\left(x+2y\right)-2\left(x+3y\right)=-6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}14x+7y-15x-5y=6\\3x+6y-2x-6y=-6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-x+2y=6\\x=-6\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=-6\\y=0\end{cases}}\)