Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
{\left( {\;3X - 2y\;} \right)^4} + {\left( {\;3y - 4z} \right)^2}\; + \left| {\;{x^2} + {y^2} + {z^2} - 1} \right|\; = 0\\
do:{\left( {\;3X - 2y\;} \right)^4} \ge 0\\
{\left( {\;3y - 4z} \right)^2}\; \ge 0\\
\left| {\;{x^2} + {y^2} + {z^2} - 1} \right|\; \ge 0\\
= > {\left( {\;3X - 2y\;} \right)^4} + {\left( {\;3y - 4z} \right)^2}\; + \left| {\;{x^2} + {y^2} + {z^2} - 1} \right|\; \ge 0
\end{array}\]

dấu = xảy ra khi và chỉ khi

\[\begin{array}{l}
= > \left\{ \begin{array}{l}
x = \frac{2}{3}y\\
z = \frac{3}{4}y\\
{\left( {\frac{2}{3}y} \right)^2} + {y^2} + {\left( {\frac{3}{4}y} \right)^2} = 1
\end{array} \right.\\
= > \left\{ \begin{array}{l}
x = \frac{2}{3}y\\
z = \frac{3}{4}y\\
{y^2} = \frac{{144}}{{289}}
\end{array} \right.\\
= > \left\{ \begin{array}{l}
\left[ \begin{array}{l}
y = \frac{{12}}{{17}}\\
x = \frac{8}{{17}}\\
z = \frac{9}{{17}}
\end{array} \right.\\
\left[ \begin{array}{l}
y = \frac{{ - 12}}{{17}}\\
x = - \frac{8}{{17}}\\
z = \frac{{ - 9}}{{17}}
\end{array} \right.
\end{array} \right.\\
vay:\left( {x;y;z} \right) = \left( {\frac{8}{{17}};\frac{{12}}{{17}};\frac{9}{{17}}} \right);\left( {\frac{{ - 8}}{{17}};\frac{{ - 12}}{{17}};\frac{{ - 9}}{{17}}} \right)
\end{array}\]

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Answer 2

\[\begin{array}{l}
{\left( {\;3X - 2y\;} \right)^4} + {\left( {\;3y - 4z} \right)^2}\; + \left| {\;{x^2} + {y^2} + {z^2} - 1} \right|\; = 0\\
do:{\left( {\;3X - 2y\;} \right)^4} \ge 0\\
{\left( {\;3y - 4z} \right)^2}\; \ge 0\\
\left| {\;{x^2} + {y^2} + {z^2} - 1} \right|\; \ge 0\\
= > {\left( {\;3X - 2y\;} \right)^4} + {\left( {\;3y - 4z} \right)^2}\; + \left| {\;{x^2} + {y^2} + {z^2} - 1} \right|\; \ge 0
\end{array}\]

dấu = xảy ra khi và chỉ khi

\[\begin{array}{l}
= > \left\{ \begin{array}{l}
x = \frac{2}{3}y\\
z = \frac{3}{4}y\\
{\left( {\frac{2}{3}y} \right)^2} + {y^2} + {\left( {\frac{3}{4}y} \right)^2} = 1
\end{array} \right.\\
= > \left\{ \begin{array}{l}
x = \frac{2}{3}y\\
z = \frac{3}{4}y\\
{y^2} = \frac{{144}}{{289}}
\end{array} \right.\\
= > \left\{ \begin{array}{l}
\left[ \begin{array}{l}
y = \frac{{12}}{{17}}\\
x = \frac{8}{{17}}\\
z = \frac{9}{{17}}
\end{array} \right.\\
\left[ \begin{array}{l}
y = \frac{{ - 12}}{{17}}\\
x = - \frac{8}{{17}}\\
z = \frac{{ - 9}}{{17}}
\end{array} \right.
\end{array} \right.\\
vay:\left( {x;y;z} \right) = \left( {\frac{8}{{17}};\frac{{12}}{{17}};\frac{9}{{17}}} \right);\left( {\frac{{ - 8}}{{17}};\frac{{ - 12}}{{17}};\frac{{ - 9}}{{17}}} \right)
\end{array}\]

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