ĐKXĐ: ...
\(x^2+6x+9+3x+10-2\sqrt{3x+10}+1=0\)
\(\Leftrightarrow\left(x+3\right)^2+\left(\sqrt{3x+10}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\\sqrt{3x+10}-1=0\end{matrix}\right.\) \(\Rightarrow x=-3\)
2/\(\frac{2}{xy}=\frac{1}{z^2}+4\)
\(\frac{1}{x}+\frac{1}{y}=2-\frac{1}{z}\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}=4+\frac{1}{z^2}-\frac{4}{z}\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+4=\frac{1}{z^2}+4-\frac{4}{z}\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}=-\frac{4}{z}\Rightarrow\frac{1}{z}=-\frac{1}{4x^2}-\frac{1}{4y^2}\)
Thay vào \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\) ta được:
\(\frac{1}{x}+\frac{1}{y}-\frac{1}{4x^2}-\frac{1}{4y^2}+2=0\)
\(\Leftrightarrow\frac{1}{4x^2}-\frac{1}{x}+1+\frac{1}{4y^2}-\frac{1}{y}+1=0\)
\(\Leftrightarrow\left(\frac{1}{2x}-1\right)^2+\left(\frac{1}{2y}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{1}{2x}-1=0\\\frac{1}{2y}-1=0\end{matrix}\right.\) \(\Leftrightarrow...\)
