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y=0 không phải nghiệm, hệ tương đương:
\(\left\{{}\begin{matrix}\left(x+1\right)^3+2=\dfrac{6}{y}\\3x+1=\dfrac{8}{y^3}\end{matrix}\right.\) \(\Rightarrow\) đặt \(\left\{{}\begin{matrix}x+1=a\\\dfrac{2}{y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^3-3b=-2\\3a-b^3=2\end{matrix}\right.\)
\(\Rightarrow a^3-b^3+3a-3b=0\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+3=0\right)\)
\(\Rightarrow a=b\) (do \(a^2+ab+b^2+3=\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}+3>0\) )
\(\Rightarrow x+1=\dfrac{2}{y}\) thay vào pt đầu:
\(y\left(\dfrac{2}{y}\right)^3+2y-6=0\Leftrightarrow2y^3-6y^2+8=0\Rightarrow\left[{}\begin{matrix}y=-1\Rightarrow x=-3\\y=2\Rightarrow x=0\end{matrix}\right.\)
Vậy hệ đã cho có 2 cặp nghiệm \(\left(x;y\right)=\left(-3;-1\right);\left(0;1\right)\)
a) thay \(x^2y^2=2y^2-1\) vào PT (2):
\(\left(xy+1\right)\left(2y-x\right)=2x\left(2y^2-1\right)\)
\(\Leftrightarrow2xy^2-x^2y+2y-x=4xy^2-2x\)
\(\Leftrightarrow2xy^2-x+x^2y-2y=0\)
\(\Leftrightarrow\left(xy-1\right)\left(2y+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}xy=1\\x=-2y\end{matrix}\right.\)...
b)
\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-6xy+2y^2=6\\x^2+2xy-2y^2=6\end{matrix}\right.\)
\(\Rightarrow3x^2-8xy+4y^2=0\)
\(\Rightarrow\left(3x-2y\right)\left(x-2y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}y=\dfrac{3}{2}x\\y=\dfrac{1}{2}x\end{matrix}\right.\)
Thế vào pt đầu...
\(\left\{{}\begin{matrix}2x^2-3xy+y^2=3\\x^2+2xy-2y^2=6\end{matrix}\right.\)\(\left(1\right)\)\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-6xy+2y^2=6\\x^2+2xy-2y^2=6\end{matrix}\right.\)
\(\Leftrightarrow3x^2-8xy+4y^2=0\)
\(\Leftrightarrow3x\left(x-2y\right)-2y\left(x-2y\right)=0\)
\(\Leftrightarrow\left(x-2y\right)\left(3x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2y\\x=\dfrac{2y}{3}\end{matrix}\right.\)
Thay vào \(\left(1\right)\) ta được:
\(\Leftrightarrow\left[{}\begin{matrix}2.\left(2y\right)^2-3.2y.y+y^2=3\\2.\left(\dfrac{2y}{3}\right)^2-3.\dfrac{2y}{3}.y+y^2=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y^2=1\\y^2=-27\left(VLý\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\end{matrix}\right.\)
Vậy ...