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\(\left\{{}\begin{matrix}\left(x+y\right)^3-3xy\left(x+y\right)=8\\x+y+2xy=2\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\) với \(a^2\ge4b\)
\(\Rightarrow\left\{{}\begin{matrix}a^3-3ab=8\\a+2b=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^3-3ab=8\\a=2-2b\end{matrix}\right.\)
\(\Rightarrow\left(2-2b\right)^3-3b\left(2-2b\right)-8=0\)
\(\Leftrightarrow2b\left(4b^2-15b+15\right)=0\)
\(\Leftrightarrow b=0\Rightarrow a=2\Rightarrow\left(x;y\right)=\left(0;2\right);\left(2;0\right)\)
\(\left\{{}\begin{matrix}mx-y=4\\x+my=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}mx=y+4\\my=-2-x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}mxy=y^2+4y\left(y\ne0\right)\\mxy=-2x-x^2\left(x\ne0\right)\end{matrix}\right.\).
Suy ra \(y^2+4y=-2x-x^2\Leftrightarrow x^2+y^2+4y+2x=0\).
\(1\))\(x^2+5x+8=3\sqrt{x^3+5x^2+7x+6}\left(1\right)\\ĐK:x\ge-\dfrac{3}{2} \\ \left(1\right)\Leftrightarrow x^2+5x+8=3\sqrt{\left(2x+3\right)\left(x^2+x+2\right)}\left(2\right)\)
Đặt \(b=\sqrt{2x+3};a=\sqrt{x^2+x+2}\)
\(\left(2\right)\Leftrightarrow\left(a-b\right)\left(a-2b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=b\\a=2b\end{matrix}\right.\)\(\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1\pm\sqrt{5}}{2}\\x=\dfrac{7\pm\sqrt{89}}{2}\end{matrix}\right.\)
4)\(ĐK:x\ge-\dfrac{1}{3}\)
\(x^2-7x+2+2\sqrt{3x+1}=0\\ \Leftrightarrow x^2-7x+6+2\sqrt{3x+1}-4=0\\ \Leftrightarrow\left(x-1\right)\left(x-6\right)+\dfrac{12\left(x-1\right)}{2\sqrt{3x+1}+4}=0\\ \Leftrightarrow\left(x-1\right)\left(x-6+\dfrac{12}{2\sqrt{3x+1}+4}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x-6+\dfrac{12}{2\sqrt{3x+1}+4}=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(x-5\right)+\dfrac{6}{\sqrt{3x+1}+2}-1=0\\ \Leftrightarrow\left(x-5\right)+\dfrac{4-\sqrt{3x+1}}{\sqrt{3x+1}+2}=0\\ \Leftrightarrow\left(x-5\right)-\dfrac{3\left(x-5\right)}{\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)}=0\\ \Leftrightarrow\left(x-5\right)\left(1-\dfrac{3}{\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\\left(1-\dfrac{3}{\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)}\right)=0\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\left(\sqrt{3x+1}+2\right)\left(4+\sqrt{3x+1}\right)=3\\ \Leftrightarrow3x+1+6\sqrt{3x+1}+8=3\\ \Leftrightarrow x+2\sqrt{3x+1}+2=0\\ \Leftrightarrow2\sqrt{3x+1}=-x-2\ge0\Leftrightarrow x\le-2\)
Vậy pt có 2 nghiệm là x=1 và x=5
(1) =>(4-x) 2>=0,(x2-2x+4>3=>(1)<0 voO LYS Vậy hệ bất phương trình vô nghiệm
ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}2\left(\frac{x^2+1}{y}\right)+2\left(x+y\right)=8\\\left(x+y\right)^2-2\left(\frac{x^2+1}{y}\right)=7\end{matrix}\right.\)
\(\Rightarrow\left(x+y\right)^2+2\left(x+y\right)=15\)
\(\Leftrightarrow\left(x+y\right)^2+2\left(x+y\right)-15=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=3\Rightarrow\frac{x^2+1}{y}=1\\x+y=-5\Rightarrow\frac{x^2+1}{y}=9\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+y=3\\\frac{x^2+1}{y}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\x^2+1=y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+x^2+1=3\\y=x^2+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+x-2=0\\y=x^2+1\end{matrix}\right.\) (casio)
TH2: \(\left\{{}\begin{matrix}x+y=-5\\\frac{x^2+1}{y}=9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=-5\\\frac{x^2+1}{9}=y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{x^2+1}{9}=-5\\y=\frac{x^2+1}{9}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+9x+46=0\\y=\frac{x^2+1}{9}\end{matrix}\right.\) (vô nghiệm)
\(\left\{{}\begin{matrix}3xy^2=x^2+2\left(1\right)\\3x^2y=y^2+2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3xy^2\left(y^2+2\right)=\left(x^2+2\right)\left(y^2+2\right)\left(2\right)\\3x^2y\left(x^2+2\right)=\left(y^2+2\right)\left(x^2+2\right)\left(3\right)\end{matrix}\right.\)
Trừ vế theo vế \(\left(2\right)\) cho \(\left(3\right)\) ta được
\(3xy^2\left(y^2+2\right)-3x^2y\left(x^2+2\right)=0\)
\(\Leftrightarrow3xy\left(y-x\right)\left(x^2+y^2+xy+2\right)=0\)
Do \(x^2+xy+y^2+2>0\forall x,y\) nên\(\left[{}\begin{matrix}x=0\\y=0\\x=y\end{matrix}\right.\)
Nếu \(x=0\Rightarrow\left\{{}\begin{matrix}0=0+2\\0=y^2+2\end{matrix}\right.\left(VN\right)\)
Nếu \(y=0\Rightarrow\left\{{}\begin{matrix}0=x^2+2\\0=0+2\end{matrix}\right.\left(VN\right)\)
Nếu \(x=y\), \(\left(1\right)\Leftrightarrow3x^3-x^2-2=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x^2+2x+2\right)=0\)
\(\Leftrightarrow x=y=1\)
Vậy hệ phương trình có nghiệm \(x=y=1\)