giải hệ CSN 1.u1+u2+u3=21 và 1/u1+1/u2+1/u3=7/12

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1. Giải thích các bước giải:

   \(\begin{array}{l}
   \left\{ \begin{array}{l}
   {u_1} + {u_2} + {u_3} = 21\\
   \frac{1}{{{u_1}}} + \frac{1}{{{u_2}}} + \frac{1}{{{u_3}}} = \frac{7}{{12}}
   \end{array} \right.\\
    \leftrightarrow \left\{ \begin{array}{l}
   {u_1} + {u_1}.q + {u_1}.{q^2} = 21\\
   \frac{1}{{{u_1}}} + \frac{1}{{{u_1}.q}} + \frac{1}{{{u_1}.{q^2}}} = \frac{7}{{12}}
   \end{array} \right.\\
    \leftrightarrow \left\{ \begin{array}{l}
   {u_1}(1 + q + {q^2}) = 21\\
   \frac{1}{{{u_1}}}\left( {1 + \frac{1}{q} + \frac{1}{{{q^2}}}} \right) = \frac{7}{{12}}
   \end{array} \right.\\
    \to (1 + q + {q^2}).\left( {1 + \frac{1}{q} + \frac{1}{{{q^2}}}} \right) = \frac{{49}}{4}\\
    \leftrightarrow (1 + q + {q^2}).\frac{{1 + {q^2} + q}}{{{q^2}}} = \frac{{49}}{4}\\
    \leftrightarrow {\left( {\frac{{1 + q + {q^2}}}{q}} \right)^2} = \frac{{49}}{4}\\
    \leftrightarrow \left[ \begin{array}{l}
   \frac{{1 + q + {q^2}}}{q} = \frac{7}{2}\\
   \frac{{1 + q + {q^2}}}{q} = \frac{{ – 7}}{2}
   \end{array} \right. \leftrightarrow \left[ \begin{array}{l}
   2 + 2q + 2{q^2} = 7q\\
   2 + 2q + 2{q^2} =  – 7q
   \end{array} \right.\\
    \leftrightarrow \left[ \begin{array}{l}
   q = 2 \to {u_1} = 3\\
   q = \frac{1}{2} \to {u_1} = 12\\
   q = \frac{{ – 9 + \sqrt {65} }}{4} \to {u_1} = \frac{{27 + 3\sqrt {65} }}{2}\\
   q = \frac{{ – 9 – \sqrt {65} }}{4} \to {u_1} = \frac{{27 – 3\sqrt {65} }}{2}
   \end{array} \right.\\

   \end{array}\)

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