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\(\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y}=2-\frac{1}{z}\\\frac{2}{xy}=4+\frac{1}{z^2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}=4+\frac{1}{z^2}-\frac{4}{z}\\\frac{2}{xy}=4+\frac{1}{z^2}\end{matrix}\right.\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}=-\frac{4}{z}\) (1)
Từ pt đầu suy ra:
\(\frac{1}{x}+\frac{1}{y}-2=-\frac{1}{z}\Rightarrow\frac{4}{x}+\frac{4}{y}-8=-\frac{4}{z}\) (2)
Thế (2) vào (1)
\(\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{x}+\frac{4}{y}-8\)
\(\Leftrightarrow\frac{1}{x^2}-\frac{4}{x}+4+\frac{1}{y^2}-\frac{4}{y}+4=0\)
\(\Leftrightarrow\left(\frac{1}{x}-2\right)^2+\left(\frac{1}{y}-2\right)^2=0\)
Bạn tự giải nốt
\(x+y+z=a\)
\(\Leftrightarrow\left(x+y+z\right)^2=a^2\)
\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=a^2\)
\(\Leftrightarrow b^2+2\left(xy+yz+zx\right)=a^2\)
\(\Leftrightarrow xy+yz+zx=\frac{a^2-b^2}{2}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\)
\(\Leftrightarrow\frac{xy+yz+zx}{xyz}=\frac{1}{c}\Leftrightarrow xyz=\left(xy+yz+zx\right)c=\frac{a^2-b^2}{2}.c\)
\(x^2+y^2+z^2=b^2\)
\(\Leftrightarrow x^2+\left(y+z\right)^2-2yz=b^2\)
\(\Leftrightarrow x^2+\left(a-x\right)^2-2\left[\frac{\left(a^2-b^2\right)c}{2x}\right]=b^2\)
\(\Leftrightarrow x^2+a^2-2ax+x^2-\frac{\left(a^2-b^2\right)c}{x}=b^2\)
\(\Leftrightarrow2x^3-2ax^2+\left(a^2-b^2\right)x-\left(a^2-b^2\right)c=0\)
\(x,y,z\) là nghiệm của phương trình trên.
~~~~~ Không chắc lắm ạ ~~~~~~
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{3}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{1}{z}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{xz+yz+z^2}=0\)
\(\Leftrightarrow\left(x+y\right)\left(xy+yz+zx+z^2\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)
Thay vào pt đầu và cuối
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x+y+z}{x\left(y+z\right)}=\frac{1}{2}\\\frac{x+y+z}{y\left(z+x\right)}=\frac{1}{3}\\\frac{x+y+z}{z\left(x+y\right)}=\frac{1}{4}\end{matrix}\right.\) lần lượt chia vế cho vế ta được hệ:
\(\left\{{}\begin{matrix}\frac{y\left(z+x\right)}{x\left(y+z\right)}=\frac{3}{2}\\\frac{z\left(x+y\right)}{x\left(y+z\right)}=2\\\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2yz=xy+3zx\\yz=2xy+xz\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2yz=xy+3zx\\3yz=6xy+3zx\end{matrix}\right.\)
\(\Rightarrow yz=5xy\Rightarrow z=5x\)
Thế vào \(yz=2xy+zx\Rightarrow5xy=2xy+5x^2\)
\(\Leftrightarrow3xy=5x^2\Rightarrow y=\frac{5x}{3}\)
Thế vào pt đầu: \(\frac{1}{x}+\frac{1}{\frac{5x}{3}+5x}=\frac{1}{2}\Rightarrow\frac{23}{20x}=\frac{1}{2}\Rightarrow x=\frac{23}{10}\)
\(\Rightarrow y=\frac{23}{6};z=\frac{23}{2}\)
\(\left\{{}\begin{matrix}y=x^2\\\frac{1}{x}=\frac{1}{y}+\frac{1}{z}\\z=xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=x^2\\\frac{1}{x}=\frac{1}{y}+\frac{1}{xy}\\z=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=x^2\\\frac{1}{x}=\frac{x}{xy}+\frac{1}{xy}=\frac{x+1}{xy}\\z=xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=x^2\\xy=x^2+x\\z=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=x^2\\x^3-x^2-x=0\\z=xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=x^2\\x\left(x^2-x-1\right)=0\\z=xy\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\left(loại\right)\\\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\end{matrix}\right.\\y=x^2\\z=xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\frac{\sqrt{5}+1}{2}\left(TM\right)\\x=\frac{1-\sqrt{5}}{2}\left(TM\right)\end{matrix}\right.\\y=x^2\\z=xy\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\frac{1+\sqrt{5}}{2}\left(\right)TM\\y=\frac{3+\sqrt{5}}{2}\left(TM\right)\\z=2+\sqrt{5}\left(TM\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=\frac{1-\sqrt{5}}{2}\\y=\frac{3-\sqrt{5}}{2}\left(TM\right)\\z=2-\sqrt{5}\left(TM\right)\end{matrix}\right.\end{matrix}\right.\)
vậy ...
bạn thử thay phương trình thứ 3 rồi sau đó thay vào phương trình 2 kết quả sẽ đúng hơn đấy