u1 + u2 + u3 + u4 + u5 = 49*(1/u1 +1/u2+…+1/u5) và u1+u3 = 35 giải hệ tìm u1 và q

0 bình luận về “u1 + u2 + u3 + u4 + u5 = 49*(1/u1 +1/u2+…+1/u5) và u1+u3 = 35 giải hệ tìm u1 và q”

1. Đáp án:

   \(\left\{ \begin{array}{l}
   q =  \pm \dfrac{1}{2}\\
   {u_1} = 28
   \end{array} \right.\)

   Giải thích các bước giải:

   Ta có:

   \(\begin{array}{l}
   \left\{ \begin{array}{l}
   {u_1} + {u_2} + {u_3} + {u_4} + {u_5} = 49.\left( {\dfrac{1}{{{u_1}}} + \dfrac{1}{{{u_2}}} + \dfrac{1}{{{u_3}}} + \dfrac{1}{{{u_4}}} + \dfrac{1}{{{u_5}}}} \right)\\
   {u_1} + {u_3} = 35
   \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
   {u_1}\left( {1 + q + {q^2} + {q^3} + {q^4}} \right) = \dfrac{{49}}{{{u_1}}}\left( {1 + \dfrac{1}{q} + \dfrac{1}{{{q^2}}} + \dfrac{1}{{{q^3}}} + \dfrac{1}{{{q^4}}}} \right)\\
   {u_1} + {u_1}.{q^2} = 35
   \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
   {u_1}.\dfrac{{{q^5} – 1}}{{q – 1}} = \dfrac{{49}}{{{u_1}}}.\dfrac{{{{\left( {\frac{1}{q}} \right)}^5} – 1}}{{\frac{1}{q} – 1}}\\
   {u_1} = \dfrac{{35}}{{1 + {q^2}}}
   \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
   \dfrac{{35}}{{1 + {q^2}}}.\dfrac{{{q^5} – 1}}{{q – 1}} = \dfrac{{49}}{{\frac{{35}}{{1 + {q^2}}}}}.\dfrac{{\frac{{1 – {q^5}}}{{{q^5}}}}}{{\frac{{1 – q}}{q}}}\\
   {u_1} = \dfrac{{35}}{{1 + {q^2}}}
   \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
   \dfrac{{35}}{{1 + {q^2}}}.\dfrac{{{q^5} – 1}}{{q – 1}} = \dfrac{{49\left( {1 + {q^2}} \right)}}{{35}}.\dfrac{{1 – {q^5}}}{{\left( {1 – q} \right).{q^4}}}\\
   {u_1} = \dfrac{{35}}{{1 + {q^2}}}
   \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
   \dfrac{{35}}{{1 + {q^2}}} = \dfrac{{49\left( {1 + {q^2}} \right)}}{{35{q^4}}}\\
   {u_1} = \dfrac{{35}}{{1 + {q^2}}}
   \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
   1225{q^4} = 49\left( {1 + 2{q^2} + {q^4}} \right)\\
   {u_1} = \dfrac{{35}}{{1 + {q^2}}}
   \end{array} \right.\\
    \Leftrightarrow \left\{ \begin{array}{l}
   1176{q^4} – 98{q^2} – 49 = 0\\
   {u_1} = \dfrac{{35}}{{1 + {q^2}}}
   \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
   \left[ \begin{array}{l}
   {q^2} =  – \dfrac{1}{6}\left( l \right)\\
   {q^2} = \dfrac{1}{4}
   \end{array} \right.\\
   {u_1} = \dfrac{{35}}{{1 + {q^2}}}
   \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
   q =  \pm \dfrac{1}{2}\\
   {u_1} = 28
   \end{array} \right.
   \end{array}\)

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