Question

Prove that approximately, when x is very small,
$\frac{3{(x+\frac{4}{9})}^{\frac{1}{2}}{(1-\frac{3}{4}{x}^2)}^{\frac{1}{3}}}{2{(1+\frac{9}{16}x)}^2}=1-\frac{307}{256}{x}^2$

Answer 1

$\frac{3{(x+\frac{4}{9})}^{\frac{1}{2}}{(1-\frac{3}{4}{x}^2)}^{\frac{1}{3}}}{2{(1+\frac{9}{16}{x}^2)}^2}$

$\Rightarrow \frac{3}{2}\times \frac{2}{3}{(1+\frac{9}{4}x)}^{1/2}{(1-\frac{3}{4}{x}^2)}^{1/3}{(1+\frac{9}{16}x)}^{-2}$

$\Rightarrow (1+\frac{9}{8}x-\frac{1}{8}\times \frac{81}{16}{x}^2)(1-\frac{1}{4}{x}^2)(1-\frac{9x}{8}+\frac{3\times 81}{256}{x}^2)$

$\Rightarrow (1+\frac{9x}{8}-\frac{81}{128}{x}^2)(1-\frac{9x}{8}+\frac{243}{256}{x}^2)(1-\frac{x^2}{4})$

$\Rightarrow (1-\frac{81x^2}{64}-\frac{81x^2}{128}+\frac{243x^2}{256})(1-\frac{x^2}{4})$

$\Rightarrow (1-\frac{243x^2}{256})(1-\frac{x^2}{4})$

$\Rightarrow 1-\frac{307}{256}{x}^2$