2 and higher powers of x can be neglected, then If x is so small that x<sup>2</sup> and higher powers of x can be neglected, then <span class="math-tex">\(\dfrac{(9+2x)^{1/2}(3+4x)}{(1-x)^{1/5}}\)</span> is approximately equal to Step-by-step Solution:
To expand the expression \( E = \frac{(9 + 2x)^{1/2}(3 + 4x)}{(1 - x)^{1/5}} \) and simplify it by neglecting higher powers of \( x \), we can proceed with the following steps:
1. Rewrite the Expression:
\[
E = (9 + 2x)^{1/2}(3 + 4x)(1 - x)^{-1/5}
\]
2. Factor Out Constants:
\[
E = \left[ 3\left(1 + \frac{2}{9}x\right)^{1/2} \right] \left[ 3\left(1 + \frac{4}{3}x\right) \right] (1 - x)^{-1/5}
\]
3. Expand Each Term Using Binomial Expansion (Neglecting Higher Powers):
\[
(1 + \frac{2}{9}x)^{1/2} \approx 1 + \frac{1}{2} \cdot \frac{2}{9}x = 1 + \frac{1}{9}x
\]
\[
(1 + \frac{4}{3}x) \approx 1 + \frac{4}{3}x
\]
\[
(1 - x)^{-1/5} \approx 1 + \frac{1}{5}x
\]
4. Multiply the Expanded Terms:
\[
E = 9 \left(1 + \frac{1}{9}x\right) \left(1 + \frac{4}{3}x\right) \left(1 + \frac{1}{5}x\right)
\]
Neglecting higher powers of \( x \):
\[
E \approx 9 \left(1 + \frac{1}{9}x + \frac{4}{3}x\right) \left(1 + \frac{1}{5}x\right)
\]
\[
E \approx 9 \left(1 + \frac{13}{9}x\right) \left(1 + \frac{1}{5}x\right)
\]
\[
E \approx 9 \left(1 + \frac{13}{9}x + \frac{1}{5}x\right)
\]
\[
E \approx 9 \left(1 + \frac{74}{45}x\right)
\]
5. Final Simplification:
\[
E \approx 9 + \frac{74}{5}x
\]
Therefore, the expanded and simplified form of \( E \) is:
\[
\boxed{9 + \frac{74}{5}x}
\]Question 30
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