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ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}xy+x+y+1=4xy\\\left(\dfrac{x}{y+1}\right)^2+\left(\dfrac{y}{x+1}\right)^2=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(y+1\right)=4xy\\\left(\dfrac{x}{y+1}\right)^2+\left(\dfrac{y}{x+1}\right)^2=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\dfrac{x}{y+1}\right)\left(\dfrac{y}{x+1}\right)=\dfrac{1}{4}\\\left(\dfrac{x}{y+1}\right)^2+\left(\dfrac{y}{x+1}\right)^2=\dfrac{1}{2}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\dfrac{x}{y+1}=u\\\dfrac{y}{x+1}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^2+v^2=\dfrac{1}{2}\\uv=\dfrac{1}{4}\end{matrix}\right.\)
\(\Rightarrow u^2-2uv+v^2=0\Leftrightarrow u=v=\pm\dfrac{1}{2}\)
TH1: \(u=v=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{y+1}=\dfrac{1}{2}\\\dfrac{y}{x+1}=\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x-y=1\\x-2y=-1\end{matrix}\right.\) \(\Leftrightarrow...\)
Th2: \(u=v=-\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{y+1}=-\dfrac{1}{2}\\\dfrac{y}{x+1}=-\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x+y=-1\\x+2y=-1\end{matrix}\right.\) \(\Leftrightarrow...\)
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\(x^3+y^3+3xy=1\Leftrightarrow\left(x+y\right)^3-1-3xy\left(x+y\right)+3xy=0\)
\(\Leftrightarrow\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)=0\)
\(\Leftrightarrow\left(x+y-1\right)\left(x^2+y^2-xy+x+y+1\right)=0\)
\(\Leftrightarrow\left(x+y-1\right)\left[\left(x-y\right)^2+\left(x+1\right)^2+\left(y+1\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y-1=0\\x=y=-1\end{matrix}\right.\)
TH1: \(x=y=-1\) thế vào pt dưới kiểm tra ko thỏa mãn
TH2: \(y=1-x\) thế vào pt dưới:
\(\sqrt{\left(4-x\right)\left(x+12\right)}=\dfrac{27}{x+3}\) (ĐKXĐ: \(-12\le x\le4;x\ne-3\))
- Với \(x< -3\) pt vô nghiệm, với \(x>-3\)
Đặt \(x+3=t>0\)
\(\Rightarrow\sqrt{\left(t+9\right)\left(7-t\right)}=\dfrac{27}{t}\Leftrightarrow64-\left(t+1\right)^2=\dfrac{27^2}{t^2}\)
\(\Leftrightarrow64=\dfrac{27^2}{t^2}+\left(t+1\right)^2=\dfrac{25^2}{t^2}+t^2+\dfrac{104}{t^2}+t+t+1\ge2\sqrt{\dfrac{25^2t^2}{t^2}}+3\sqrt[3]{\dfrac{104t^2}{t^2}}+1>65\) (vô lý)
Vậy hệ vô nghiệm
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\(\Leftrightarrow xy\left(x+y\right)+\left(x+y\right)^2-2xy=4\)
\(\Leftrightarrow xy\left(x+y-2\right)+\left(x+y-2\right)\left(x+y+2\right)=0\)
\(\Leftrightarrow\left(x+y-2\right)\left(x+y+xy+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y-2=0\left(1\right)\\x+y+xy+2=0\left(2\right)\end{matrix}\right.\)
Xét (1) \(\Leftrightarrow y=2-x\) thay vào pt đầu: ....
Xét (2): kết hợp với pt đầu ta được:
\(\left\{{}\begin{matrix}x+y+xy+2=0\\\left(x+y\right)^3-3xy\left(x+y\right)-3xy=-1\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\) với \(a^2\ge4b\)
\(\Rightarrow\left\{{}\begin{matrix}a+b+2=0\\a^3-3ab-3b=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b+2=0\\\left(a+1\right)\left(a^2-a+1\right)-3b\left(a+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b+2=0\\\left(a+1\right)\left(a^2-a+1-3b\right)=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
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ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\frac{x}{y+1}\right)^2+\left(\frac{y}{x+1}\right)^2=\frac{1}{2}\\4xy=xy+x+y+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\frac{x}{y+1}\right)^2+\left(\frac{y}{x+1}\right)^2=\frac{1}{2}\\4xy=\left(x+1\right)\left(y+1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\frac{x}{y+1}\right)^2+\left(\frac{y}{x+1}\right)^2=\frac{1}{2}\\\left(\frac{x}{y+1}\right)\left(\frac{y}{x+1}\right)=\frac{1}{4}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\frac{x}{y+1}=a\\\frac{y}{x+1}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2+b^2=\frac{1}{2}\\ab=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow a=b=\pm\frac{1}{2}\)
TH1: \(a=b=\frac{1}{2}\Rightarrow\left\{{}\begin{matrix}\frac{x}{y+1}=\frac{1}{2}\\\frac{y}{x+1}=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=y+1\\2y=x+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x-y=1\\x-2y=-1\end{matrix}\right.\) (bấm máy)
TH2: \(a=b=-\frac{1}{2}\) tương tự
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2(x+y)+16=3xy
⇔6(x+y)+48=9xy
⇔6x−9xy−4+6y+52=0
⇔3x(2−3y)−2(2−3y)=−52
⇔(2−3y)(3x−2)=−52
Em tham khảo thêm: Câu hỏi của Tiểu thư họ Vũ - Toán lớp 9 - Học toán với OnlineMath