Cho x,y,z $\neq$ 0
thỏa mãn
$\frac{1}{x}$ + $\frac{1}{y}$ + $\frac{1}{z}$ =2
Và $\frac{2}{xy}$ – $\frac{1}{z^2}$ =4
Tính P=(x+2y+z)^2019
Cho x,y,z $\neq$ 0
thỏa mãn
$\frac{1}{x}$ + $\frac{1}{y}$ + $\frac{1}{z}$ =2
Và $\frac{2}{xy}$ – $\frac{1}{z^2}$ =4
Tính P=(x+2y+z)^2019
Đáp án:
\[P = 1\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 2\\
\frac{2}{{xy}} – \frac{1}{{{z^2}}} = 4
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\frac{1}{z} = 2 – \frac{1}{x} – \frac{1}{y}\\
\frac{1}{{{z^2}}} = \frac{2}{{xy}} – 4
\end{array} \right.\\
\Leftrightarrow {\left( {2 – \frac{1}{x} – \frac{1}{y}} \right)^2} = \frac{2}{{xy}} – 4\\
\Leftrightarrow 4 + \frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} – \frac{4}{x} – \frac{4}{y} + \frac{2}{{xy}} = \frac{2}{{xy}} – 4\\
\Leftrightarrow \frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} – \frac{4}{x} – \frac{4}{y} + 8 = 0\\
\Leftrightarrow \left( {\frac{1}{{{x^2}}} – \frac{4}{x} + 4} \right) + \left( {\frac{1}{{{y^2}}} – \frac{4}{y} + 4} \right) = 0\\
\Leftrightarrow {\left( {\frac{1}{x} – 2} \right)^2} + {\left( {\frac{1}{y} – 2} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
\frac{1}{x} – 2 = 0\\
\frac{1}{y} – 2 = 0
\end{array} \right. \Leftrightarrow x = y = \frac{1}{2} \Rightarrow z = – \frac{1}{2}\\
\Rightarrow P = {\left( {x + 2y + z} \right)^{2019}} = {\left( {\frac{1}{2} + 1 – \frac{1}{2}} \right)^{2019}} = 1
\end{array}\)