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

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

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

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

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

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

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

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

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

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

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 

\(\sqrt{x+1}-\sqrt{y+1}+\sqrt{9-y}-\sqrt{9-x}=0\)Liên hợp có x-y=0

thay vào PT đầu

\(\sqrt{x+1}+\sqrt{9-x}=4\)

BP

\(\sqrt{\left(x+1\right)\left(9-x\right)}=3\) 

(x+1)((9-x)=9=> x=0 hoạc x=8

(xy)=(0,0);(8,8)

Hình như sai bạn ak

trừ cho nhau là xong

Nói nghe có vẻ dễ ha Trần Hữu Ngọc Minh