解決(x*x^2-x^2-14x-12)div(x^2-x-12) |Microsoft數學求解器

評估

\frac{\left(x+1\right)\left(x^{2}-2x-12\right)}{\left(x-4\right)\left(x+3\right)}

簡短的解題步驟

(x \times { x }^{ 2 } - { x }^{ 2 } -14x-12) \div ( { x }^{ 2 } -x-12)

計算有相同底數之乘冪數相乘的方法: 將指數相加。1 加 2 得到 3。

\frac{x^{3}-x^{2}-14x-12}{x^{2}-x-12}

展開

\frac{x^{3}-x^{2}-14x-12}{x^{2}-x-12}

簡短的解題步驟

(x \times { x }^{ 2 } - { x }^{ 2 } -14x-12) \div ( { x }^{ 2 } -x-12)

計算有相同底數之乘冪數相乘的方法: 將指數相加。1 加 2 得到 3。

\frac{x^{3}-x^{2}-14x-12}{x^{2}-x-12}

圖表

測驗

Polynomial

(x \times { x }^{ 2 } - { x }^{ 2 } -14x-12) \div ( { x }^{ 2 } -x-12)

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Torsion and curvature of the curve X(t) = (at, bt^2, ct^3)

https://math.stackexchange.com/q/1185846

Those calculations with space curves tend to be tedious; therefore using a computer algebra system is not quite a luxury. Here goes: \vec{r}(t) = \left[ \begin{matrix} a t \\ b t^2 \\ c t^3 \end{matrix} \right] \quad \Longrightarrow \\ \vec{\iota} = \, \stackrel{.}{\vec{r}} \, = \frac{d\vec{r}}{ds} = \frac{d\vec{r}}{dt}\frac{dt}{ds} = \left[ \begin{matrix} a \\ 2 b t \\ 3 c t^2 \end{matrix} \right] / \sqrt{a^2 + 4 b^2 t^2 + 9 c^2 t^4} \quad \Longrightarrow \\ \stackrel{..}{\vec{r}} \, = \frac{d^2 \vec{r}}{ds^2} = \, \stackrel{.}{\vec{\iota}} \, = \frac{d\vec{\iota}}{ds} = \frac{d\vec{\iota}}{dt}\frac{dt}{ds} = \left[ \begin{matrix} -2at(2b^2+9c^2t^2)\\2b(a^2-9c^2t^4)\\6ct(a^2+2b^2t^2) \end{matrix} \right]/(a^2+4b^2t^2+9c^2t^4)^2 \quad \Longrightarrow \\ \stackrel{...}{\vec{r}} \, = \, \stackrel{..}{\vec{\iota}} \, = \frac{d\stackrel{.}{\vec{\iota}}}{dt} \frac{dt}{ds} = \left[ \begin{matrix} 2a(2a^2b^2-24t^2b^4-162t^4b^2c^2+27a^2c^2t^2-405t^6c^4) \\ -8bt(27a^2c^2t^2-81t^6c^4+4a^2b^2) \\ 6c(a^4-6a^2b^2t^2-63a^2c^2t^4-8b^4t^4-90b^2t^6c^2) \end{matrix} \right] / (a^2+4b^2t^2+9c^2t^4)^{7/2} ...

How prove this matrix equation AXB=C has a solution

https://math.stackexchange.com/q/585419

I will assume that your matrices are over a regular ring (maybe even over a field). Then each of these is regular. By this I mean for A there is matrix A^- such that AA^-A=A. Then the first two ...

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\frac{x^{3}-x^{2}-14x-12}{x^{2}-x-12}

計算有相同底數之乘冪數相乘的方法: 將指數相加。1 加 2 得到 3。

\frac{x^{3}-x^{2}-14x-12}{x^{2}-x-12}

計算有相同底數之乘冪數相乘的方法: 將指數相加。1 加 2 得到 3。

示例

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三角學

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聯立方程

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微分

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積分

\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x

限制

\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}

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