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\(
\begin{aligned}
x_{1}^{1} &= \frac{1}{7}(3 - 0 - 0) = \frac{3}{7} \\
x_{2}^{1} &= \frac{1}{8}(-2 - (2)x_{1}^{1}) = \frac{-1}{7} \\
x_{3}^{1} &= \frac{1}{5}(5 - (-1)x_{1}^{1} - (0)x_{2}^{1}) = \frac{38}{35} \\
x_{4}^{1} &= \frac{1}{4}(4 - (0)x_{1}^{1} - (2)x_{2}^{1} - (-1)x_{3}^{1}) = \frac{29}{20}
\end{aligned}

\)
\(\begin{equation} \alpha^2 \end{equation}\)



\(
\begin{array}{l}
f(x) = (x^3  + 1)(x - 1)Q(x) + a(x^3  + 1) - x^2  - 3x - 3 \\
x - 1 = 0 \\
a = 5 \\
y - a \\
\end{array}
\)

[ 本帖最後由 eyeready 於 2017-4-15 18:10 編輯 ]

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