\(\left\{{}\begin{matrix}\left(x+y\right).\left(x^2+y^2\right)=15\\\left(x-y\right).\left(x^2-y^2\right)=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right).\left(x^2+y^2\right)=15\\\left(x+y\right).\left(x-y\right)^2=3\end{matrix}\right.\)
\(\Rightarrow\frac{x^2+y^2}{\left(x-y\right)^2}=\frac{15}{3}=5\left(\text{vì }x+y\ne0\right)\)
\(\Rightarrow5.\left(x-y\right)^2-x^2-y^2=0\)
\(\Rightarrow\left[\left(x-y\right)^2-x^2\right]+\left[\left(2x-2y\right)^2-y^2\right]=0\)
\(\Rightarrow\left(x-y-x\right).\left(x-y+x\right)+\left(2x-2y-y\right).\left(2x-2y+y\right)=0\)
\(\Rightarrow-y.\left(2x-y\right)+\left(2x-3y\right).\left(2x-y\right)=0\)
\(\Rightarrow\left(2x-y\right).\left(2x-3y-y\right)=0\)
\(\Rightarrow\left(2x-y\right).\left(2x-4y\right)=0\Rightarrow\left[{}\begin{matrix}2x=y\\x=2y\end{matrix}\right.\)
Thay vào giải tiếp
Lười nghĩ quá :v cứ nhân tung tóe ra vậy
a) \(pt\Leftrightarrow\left(x+y\right)\left(x^2+y^2\right)=5\left(x-y\right)\left(x^2-y^2\right)\)
\(\Leftrightarrow4x^3+4y^3+6x^2y+6xy^2=0\)
\(\Leftrightarrow2x^3+2y^3+3x^2y+3xy^2=0\)
\(\Leftrightarrow2\left(x+y\right)\left(x^2-xy+y^2\right)+3xy\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(2x^2+xy+2y^2\right)=0\)
Vì \(2x^2+xy+2y^2>0\forall x;y\)
\(\Rightarrow x=-y\)
Thế x vào tìm y
a) \(\left\{{}\begin{matrix}x^3+y^3+x^2y+xy^2=15\\x^3+y^3-x^2y-xy^2=3\end{matrix}\right.\)⇒ \(\left\{{}\begin{matrix}x^3+y^3=9\\x^2y+xy^2=6\end{matrix}\right.\)
⇒ \(x^3+y^3+3x^2y+3xy^2=27\)
⇒ \(x+y=3\Rightarrow x^2+y^2=5\)(1)
Thay \(y=3-x\) vào (1) , ta được:
\(2x^2-6x+4=0\Rightarrow\left(x-1\right)\left(x-2\right)=0\)
⇒ \(\left[{}\begin{matrix}x=1;y=2\\x=2;y=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2+y^2+1=2\left(x+y\right)\\y\left(2x-y\right)=2y+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-2x+1+y^2-2y+1-1=0\\y^2-2xy+2y+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2+\left(y-1\right)^2-1=0\\\left(y-1\right)^2-2xy=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(y-1\right)^2=1-\left(x-1\right)^2\\\left(y-1\right)^2=2xy\end{matrix}\right.\)
\(\Rightarrow1-\left(x-1\right)^2=2xy\)
\(\Leftrightarrow x^2-2x+2xy=0\)
\(\Leftrightarrow x\left(x+2y-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+2y=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2-2y\end{matrix}\right.\)
Thế vào tìm y nốt nhé