Toán Lớp 11: (1+x)(1+x^2)(1+x^4)=1+x^7 giải pt
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Toán Lớp 11: (1+x)(1+x^2)(1+x^4)=1+x^7 giải pt
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\left( {1 + x} \right)\left( {1 + {x^2}} \right)\left( {1 + {x^4}} \right) = 1 + {x^7}\\
\Leftrightarrow \left( {1 + x} \right)\left( {1 + {x^2}} \right)\left( {1 + {x^4}} \right) = \left( {x + 1} \right)\left( {{x^6} – {x^5} + {x^4} – {x^3} + {x^2} – x + 1} \right)\\
\Leftrightarrow \left( {x + 1} \right)\left( {{x^6} – {x^5} + {x^4} – {x^3} + {x^2} – x + 1 – \left( {{x^2} + 1} \right)\left( {{x^4} + 1} \right)} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {{x^6} – {x^5} + {x^4} – {x^3} + {x^2} – x – 1 – {x^6} – {x^2} – {x^4} – 1} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {{x^5} – {x^3} – x} \right) = 0\\
\Leftrightarrow \left( {x + 1} \right)x\left( {{x^4} – {x^2} – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = – 1\\
x = 0\\
{x^4} – {x^2} – 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = – 1\\
x = 0\\
{x^2} = \dfrac{{1 + \sqrt 5 }}{2}\\
{x^2} = \dfrac{{1 – \sqrt 5 }}{2}(L)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = – 1\\
x = 0\\
x = \pm \sqrt {\dfrac{{1 + \sqrt 5 }}{2}}
\end{array} \right.
\end{array}$