Giải thích các bước giải:
Câu 2:
1)
ĐK: $x\ne 0$
Ta có:
$\begin{array}{l}
x + \dfrac{1}{x} = {x^2} + \dfrac{1}{{{x^2}}}\\
\Leftrightarrow {\left( {x + \dfrac{1}{x}} \right)^2} - 2 - \left( {x + \dfrac{1}{x}} \right) = 0\\
\Leftrightarrow {\left( {x + \dfrac{1}{x}} \right)^2} - \left( {x + \dfrac{1}{x}} \right) - 2 = 0\\
\Leftrightarrow \left( {x + \dfrac{1}{x} + 1} \right)\left( {x + \dfrac{1}{x} - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{1}{x} + 1 = 0\\
x + \dfrac{1}{x} - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{x^2} + x + 1 = 0\\
\left( {vn,do:{x^2} + x + 1 = {{\left( {x + \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4} \ge \dfrac{3}{4} > 0,\dforall x} \right)\\
{x^2} - 2x + 1 = 0
\end{array} \right.\\
\Leftrightarrow {x^2} - 2x + 1 = 0\\
\Leftrightarrow {\left( {x - 1} \right)^2} = 0\\
\Leftrightarrow x = 1
\end{array}$
Vậy phương trình có tập nghiệm $S=\left{1\right}$
2) ĐK: $x,y,z\ne 0$
Ta có:
$\begin{array}{l}
\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 2\\
\Leftrightarrow \dfrac{1}{x} + \dfrac{1}{y} = 2 - \dfrac{1}{z}\\
\Rightarrow {\left( {\dfrac{1}{x} + \dfrac{1}{y}} \right)^2} = {\left( {2 - \dfrac{1}{z}} \right)^2}\\
\Leftrightarrow \dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}} + \dfrac{2}{{xy}} = 4 - \dfrac{4}{z} + \dfrac{1}{{{z^2}}}\\
\Leftrightarrow \dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}} = \dfrac{{ - 4}}{z}\\
\Leftrightarrow \dfrac{1}{{{x^2}}} + \dfrac{1}{{{y^2}}} - 4\left( {\dfrac{1}{x} + \dfrac{1}{y}} \right) + 8 = 0\left( {do:\dfrac{1}{x} + \dfrac{1}{y} = 2 - \dfrac{1}{z}} \right)\\
\Leftrightarrow \left( {\dfrac{1}{{{x^2}}} - 4.\dfrac{1}{x} + 4} \right) + \left( {\dfrac{1}{{{y^2}}} - 4.\dfrac{1}{y} + 4} \right) = 0\\
\Leftrightarrow {\left( {\dfrac{1}{x} - 2} \right)^2} + {\left( {\dfrac{1}{y} - 2} \right)^2} = 0\\
\Leftrightarrow {\left( {\dfrac{1}{x} - 2} \right)^2} = {\left( {\dfrac{1}{y} - 2} \right)^2} = 0\\
\left( {Do:{{\left( {\dfrac{1}{x} - 2} \right)}^2} + {{\left( {\dfrac{1}{y} - 2} \right)}^2} \ge 0,\dforall x,y \ne 0} \right)\\
\Leftrightarrow \dfrac{1}{x} - 2 = \dfrac{1}{y} - 2 = 0\\
\Leftrightarrow x = y = \dfrac{1}{2}
\end{array}$
Khi đó:
$\dfrac{1}{z} = 2 - \dfrac{1}{x} - \dfrac{1}{y} = - 2 \Rightarrow z = \dfrac{{ - 1}}{2}$
Vậy $\left( {x;y;z} \right) = \left( {\dfrac{1}{2};\dfrac{1}{2};\dfrac{{ - 1}}{2}} \right)$
