If x is small (1-x)^-5/2+(16+8x)^1/2÷(1+x)^-1/2(2+x)^2

We are asked to simplify or analyze the expression:

\[
(1 - x)^{-5/2} + \frac{(16 + 8x)^{1/2}}{(1 + x)^{-1/2}} \cdot (2 + x)^2
\]

Let’s carefully simplify each part, considering \(x\) is small. When \(x\) is small, we can make use of **approximations** based on the binomial series expansion. 

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### Step-by-Step Solution

#### 1. Simplifying \((1 - x)^{-5/2}\) for small \(x\):
Use the **binomial expansion** for small \(x\):
\[
(1 - x)^{n} \approx 1 + nx + \frac{n(n-1)}{2}x^2 + \dots
\]
Here \(n = -5/2\), so:
\[
(1 - x)^{-5/2} \approx 1 + \left(-\frac{5}{2}\right)(-x) + \frac{\left(-\frac{5}{2}\right)\left(-\frac{7}{2}\right)}{2}x^2 + \dots
\]
\[
(1 - x)^{-5/2} \approx 1 + \frac{5}{2}x + \frac{35}{8}x^2 + \dots
\]

#### 2. Simplifying \((16 + 8x)^{1/2}\) for small \(x\):
We can factor out 16:
\[
(16 + 8x)^{1/2} = \sqrt{16(1 + \frac{1}{2}x)} = 4 \cdot (1 + \frac{1}{2}x)^{1/2}
\]
Now, use the binomial expansion again:
\[
(1 + \frac{1}{2}x)^{1/2} \approx 1 + \frac{1}{2} \cdot \frac{1}{2}x = 1 + \frac{1}{4}x
\]
Thus:
\[
(16 + 8x)^{1/2} \approx 4 \left(1 + \frac{1}{4}x\right) = 4 + x
\]

#### 3. Simplifying \((1 + x)^{-1/2}\) for small \(x\):
Using the binomial expansion for \((1 + x)^{-1/2}\):
\[
(1 + x)^{-1/2} \approx 1 - \frac{1}{2}x + \dots
\]

#### 4. Simplifying the product with \((2 + x)^2\):
Expand \((2 + x)^2\):
\[
(2 + x)^2 = 4 + 4x + x^2
\]

#### 5. Putting it all together:
Now, the full second term becomes:
\[
\frac{(4 + x)}{1 - \frac{1}{2}x} \cdot (4 + 4x + x^2)
\]
For small \(x\), use the approximation \(\frac{1}{1 - \frac{1}{2}x} \approx 1 + \frac{1}{2}x\):
\[
\frac{4 + x}{1 - \frac{1}{2}x} \approx (4 + x)(1 + \frac{1}{2}x) = 4 + x + 2x + \frac{1}{2}x^2 = 4 + 3x + \frac{1}{2}x^2
\]
Now multiply by \(4 + 4x + x^2\):
\[
(4 + 3x + \frac{1}{2}x^2)(4 + 4x + x^2)
\]
Expanding:
\[
16 + 16x + 4x^2 + 12x + 12x^2 + \frac{3}{2}x^3 + 2x^2 + 2x^3 + \frac{1}{2}x^4
\]
Simplify the terms:
\[
16 + 28x + 18x^2 + \frac{7}{2}x^3 + \frac{1}{2}x^4
\]

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### Final Expression:
Now, combining the two terms, we get:
\[
(1 - x)^{-5/2} + \left(16 + 28x + 18x^2 + \frac{7}{2}x^3 + \frac{1}{2}x^4\right)
\]
Using the earlier approximation for \((1 - x)^{-5/2}\):
\[
1 + \frac{5}{2}x + \frac{35}{8}x^2 + \dots + 16 + 28x + 18x^2 + \frac{7}{2}x^3 + \frac{1}{2}x^4
\]

Simplify the combined terms:
\[
17 + \left(\frac{5}{2} + 28\right)x + \left(\frac{35}{8} + 18\right)x^2 + \dots
\]
\[
17 + \frac{61}{2}x + \frac{179}{8}x^2 + \dots
\]

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### Conclusion:
The expression simplifies to:
\[
17 + \frac{61}{2}x + \frac{179}{8}x^2 + \dots
\]

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Do you want further details or have any questions?

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### 5 Related Questions:
1. What is the binomial expansion, and when is it used?
2. How does the approximation of small values work in calculus?
3. What are higher-order terms in Taylor series expansions?
4. How can we simplify complex algebraic fractions?
5. What are common uses of fractional exponents in physics and engineering?

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### Tip:
When \(x\) is small, binomial approximations are extremely useful for quick calculations. Remember to identify when higher-order terms are insignificant to avoid unnecessary complexity!

Simplify the complex expression involving binomial expansion and small x approximations: (1 - x)^(-5/2) + (16 + 8x)^(1/2) / (1 + x)^(-1/2) * (2 + x)^2. This step-by-step solution applies key algebraic and series expansion techniques, suitable for students in Grades 11-12.

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