Đặt
\(a=\sqrt{x+y}\rightarrow a^2=x+y,a\ge0; \)
\(b=\sqrt{x-2y}\rightarrow b^2=x-2y,b\ge0\)
Phương trình (1) có :
x3+3x2y+4y3=4(x-2y)2
\(\Leftrightarrow\)x3+y3-3x2y+3y3=4(x-2y)2
\(\Leftrightarrow\)(x+y)(x2-xy+y2)-3y(x2-y2)=4(x-2y)2
\(\Leftrightarrow\)(x+y)(x2-xy+y2)-3y(x+y)(x-y)=4(x-2y)2
\(\Leftrightarrow\)(x+y)(x2-xy+y2-3y(x-y))=4(x-2y)2
\(\Leftrightarrow\)(x+y)(x2-4xy+4y2)=4(x-2y)2
\(\Leftrightarrow\)(x+y)(x-2y)2=4(x-2y)2
\(\Leftrightarrow\)a2b4=4b4
\(\Leftrightarrow\)b4(a2-4)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}b=0\\a-2=0\\a+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}b=0\\a=2\\a=-2\left(lo\text{ại}\right)\end{matrix}\right.\)
hệ pt tương đuơng
\(\left\{{}\begin{matrix}a+b=3\\b=0\end{matrix}\right.\)hoặc\(\left\{{}\begin{matrix}a+b=3\\a=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=0\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)
Với \(\left\{{}\begin{matrix}a=3\\b=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=9\\x-2y=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=6\\y=3\end{matrix}\right.\)
Với \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\x-2y=1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)