Question

Giải hệ: \(\left\{{}\begin{matrix}x^2+y^2-xy=1\\x+ỹ^2=2y^3\end{matrix}\right.\)

Answer 1

Đề lỗi VNI kìa

Answer 2

gõ lại đề đi e

Answer 3

\(\left\{{}\begin{matrix}x^2+y^2-xy=1\\x+yx^2=2y^3\end{matrix}\right.\)

Answer 4

\(HPT\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=1+xy\left(1\right)\\x\left(1+xy\right)=2y^3\left(2\right)\end{matrix}\right.\)

Thế \(\left(1\right)\) vào \(\left(2\right)\)

\(\Leftrightarrow x\left(x^2+y^2\right)=2y^3\\ \Leftrightarrow x^3+xy^2-2y^3=0\\ \Leftrightarrow x^3-xy^2+2xy^2-2y^3=0\\ \Leftrightarrow x\left(x^2-y^2\right)+2y^2\left(x-y\right)=0\\ \Leftrightarrow x\left(x-y\right)\left(x+y\right)+2y^2\left(x-y\right)=0\\ \Leftrightarrow\left(x-y\right)\left(x^2+xy+2y^2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=y\\\left(x+\dfrac{1}{2}y\right)^2+\dfrac{7}{4}y^2=0\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=y\)

Thay vào pt \(\left(1\right)\Leftrightarrow x^2+x^2-x^2=1\Leftrightarrow x^2=1\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\)

Vậy hpt có nghiệm \(\left(x;y\right)=\left\{\left(1;1\right);\left(-1;-1\right)\right\}\)