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Câu a pt đầu là \(x^2+2xy^2=3\) hay \(x^3+2xy^2=3\) vậy nhỉ? Nhìn \(x^2\) chẳng hợp lý chút nào
b. \(\Leftrightarrow\left\{{}\begin{matrix}x^2\left(xy+1\right)-y\left(xy+1\right)+xy+1=2\\\left(x^4+y^2-2x^2y\right)+xy+1=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-y\right)\left(xy+1\right)+xy+1=2\\\left(x^2-y\right)^2+xy+1=2\end{matrix}\right.\)
Trừ vế cho vế:
\(\left(x^2-y\right)\left(xy+1\right)-\left(x^2-y\right)^2=0\)
\(\Leftrightarrow\left(x^2-y\right)\left(xy+1-x^2+y\right)=0\)
\(\Leftrightarrow\left(x^2-y\right)\left[y\left(x+1\right)+\left(x+1\right)\left(1-x\right)\right]=0\)
\(\Leftrightarrow\left(x^2-y\right)\left(x+1\right)\left(y+1-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=x^2\\x=-1\\y=x-1\end{matrix}\right.\)
- Với \(y=x^2\) thế xuống pt dưới:
\(x^4+x^4-x^3\left(2x-1\right)=1\Leftrightarrow x^3=1\Leftrightarrow...\)
....
Hai trường hợp còn lại bạn tự thế tương tự
\(\Leftrightarrow\left\{{}\begin{matrix}x^2\left(xy+1\right)-y\left(xy+1\right)+xy+1=2\\\left(x^2-y\right)^2+xy+1=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-y+1\right)\left(xy+1\right)=2\\\left(x^2-y\right)^2+xy+1=2\end{matrix}\right.\)
\(\Rightarrow\left(x^2-y+1\right)\left(xy+1\right)-\left(x^2-y\right)^2-\left(xy+1\right)=0\)
\(\Leftrightarrow\left(xy+1\right)\left(x^2-y\right)-\left(x^2-y\right)^2=0\)
\(\Leftrightarrow\left(x^2-y\right)\left(xy+1-x^2+y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}y=x^2\\xy+1=x^2-y\end{matrix}\right.\) thay xuống pt dưới:
- Với \(y=x^2\) thay xuống pt dưới \(\Rightarrow x^3=1\)
- Với \(xy+1=x^2-y\) thay xuống dưới:
\(\left\{{}\begin{matrix}xy+1=x^2-y\\2\left(xy+1\right)=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}xy+1=x^2-y\\xy=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0;y=-1\\y=0;x^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x^2-2x\right)-\left(y^2-4y\right)=1\\\left(x^2-2x\right)^2+2=y\left(x-2\right)x\left(y-4\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x^2-2x\right)-\left(y^2-4y\right)=1\\\left(x^2-2x\right)^2+2=\left(x^2-2x\right)\left(y^2-4y\right)\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2-2x=u\\y^2-4y=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2u-v=1\\u^2+2=uv\end{matrix}\right.\) \(\Rightarrow u^2+2=u\left(2u-1\right)\)
\(\Leftrightarrow u^2-u-2=0\Leftrightarrow...\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2\left(xy+1\right)-y\left(xy+1\right)+xy+1=2\\\left(x^2-y^{ }\right)^2+xy+1=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-y+1\right)\left(xy+1\right)=2\\\left(x^2-y\right)^2+xy=2\end{matrix}\right.\)
\(\Rightarrow\left(x^2-y+1\right)\left(xy+1\right)-\left(x^2-y\right)^2-\left(xy+1\right)=0\)
\(\Leftrightarrow\left(xy+1\right)\left(x^2-y\right)-\left(x^2-y\right)^2=0\)
\(\Leftrightarrow\left(x^2-y\right)\left(xy+1-x^2+y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}y=x^2\\xy+1=x^2-y\end{matrix}\right.\)thay PT xuống dưới
Với \(y=x^2\) thay xuống PT dưới \(\Rightarrow x^3=1\)
Với \(xy+1=x^2-y\) thay xuống dưới:
\(\left\{{}\begin{matrix}xy+1=x^2-y\\2\left(xy+1\right)=2\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}xy+1=x^2-y\\xy=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0;y=-1\\y=0;x^2=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+x^3y-xy^2+xy-y=1\left(1\right)\\x^4+y^2-xy\left(2x-1\right)=1\left(2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2-y\right)+xy\left(x^2-y\right)+xy=1\\\left(x^2-y\right)^2+xy=1\end{matrix}\right.\)
Đặt: \(\left\{{}\begin{matrix}a=x^2-y\\b=xy\end{matrix}\right.\)
Ta có hệ phương trình mới:
\(\left\{{}\begin{matrix}a+ab+b=1\\a^2+b=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a^3+a^2-2a=0\\b=1-a^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\left(a^2+a-2\right)=0\\b=1-a^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=0;1;-2\\b=1;0;-3\end{matrix}\right.\)
Với: \(\left(a,b\right)=\left(0;1\right)\) ta có hệ: \(\left\{{}\begin{matrix}x^2-y=0\\xy=1\end{matrix}\right.\Leftrightarrow x=y=1\)
Tương tự như trên .....
Vậy hệ pt có nghiệm \(\left(x,y\right)=\left\{\left(1;1\right);\left(0;-1\right);\left(-1;0\right);\left(1;0\right);\left(-1;3\right)\right\}\)