Solve (4u^2+6/u^2)^8 | Microsoft Math Solver

Evaluate

\frac{256\left(2u^{4}+3\right)^{8}}{u^{16}}

Expand

65536u^{16}+786432u^{12}+4128768u^{8}+12386304u^{4}+23224320+\frac{27869184}{u^{4}}+\frac{20901888}{u^{8}}+\frac{8957952}{u^{12}}+\frac{1679616}{u^{16}}

Quiz

Polynomial

5 problems similar to:

( 4 u ^ { 2 } + \frac { 6 } { u ^ { 2 } } ) ^ { 8 }

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u^2+3u/u^2-9

https://www.tiger-algebra.com/drill/u~2_3u/u~2-9/

u2+3u/u2-9 Final result : u3 - 9u + 3 ——————————— u Step by step solution : Step 1 : u Simplify —— u2 Dividing exponential expressions : 1.1 u1 divided by u2 = u(1 - 2) = u(-1) = 1/u1 = 1/u ...

How do you simplify \displaystyle\frac{{{5}{u}{v}^{{6}}}}{{{u}^{{2}}{v}^{{2}}}} ?

https://socratic.org/questions/how-do-you-simplify-5uv-6-u-2v-2

\displaystyle\frac{{{5}{u}{v}^{{6}}}}{{{u}^{{2}}{v}^{{2}}}}=\frac{{{5}{v}^{{4}}}}{{u}} Explanation: \displaystyle\frac{{{5}{u}{v}^{{6}}}}{{{u}^{{2}}{v}^{{2}}}} Apply exponent quotient ...

Image of Upper Half Disc under w = 1/z

https://math.stackexchange.com/questions/1647698/image-of-upper-half-disc-under-w-1-z

Your answer is true but you could go faster with polar form. If z = re^{i\theta} with r<1 and \theta \in ]0,\pi[ then \frac{1}{z} = \frac{1}{r} e^{-i\theta} where \frac{1}{r} > 1 and -\theta \in ]-\pi,0[.

Which rational functions are derivatives of rational functions?

https://math.stackexchange.com/questions/297522/which-rational-functions-are-derivatives-of-rational-functions

Examine the poles of your function (in the complex plane). If all residues are zero, you are in good shape.

integral of \frac {\sqrt {x+4}}x

https://math.stackexchange.com/q/1780347

Let I be the integral I=\int \frac{\sqrt{x+4}}{x}\,dx Now, substitute u=\sqrt{x+4} to that x=u^2-4 and dx=2u\,du. Then, I becomes \begin{align} I&=2\int \frac{u^2}{u^2-4}\,du\\\\ &=2\int \left(\frac{u^2-4+4}{u^2-4}\right)\,du\\\\ &=2u+2\int\left(\frac{1}{u-2}-\frac{1}{u+2}\right)\,du\\\\ &=2u+2\log\left|\frac{u-2}{u+2}\right|+C\\\\ &=2\sqrt{x+4}+2\log\left|\frac{\sqrt{x+4}-2}{\sqrt{x+4}+2}\right|+C \end{align}

Transformations of Normal

https://math.stackexchange.com/questions/34007/transformations-of-normal

\frac{V+W}{\sigma^2} \sim N(0, 2) and the ratio of two iid standard normal is Cauchy(0,1). So \frac{U}{(V+W)/\sqrt{2}} \sim \text{Cauchy}(0,1). You can then derive that \frac{U}{V+W} \sim \text{Cauchy}(0, \sqrt{2}) ...

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