A) 3 |
B) \( \Large \frac{3}{2} \) |
C) \( \Large \frac{2}{3} \) |
D) \( \Large \frac{1}{3} \) |
D) \( \Large \frac{1}{3} \) |
To find :
\( \Large \sqrt{\frac{ \left(0.064-0.008\right) \left(0.16-0.04\right) }{ \left(0.16+0.08+0.04\right) \left(0.4+0.2\right)^{3} }} \)
Let x = 0.4 ; y = 0.2
\( \Large \sqrt\frac{ \left(x^{3}-y^{3}\right) \left(x^{2}-y^{2}\right) }{ \left(x^{2}+xy+y^{2}\right) \left(x+y\right)^{3} } \)
since \( \Large \left(x^{3}-y^{3}\right) = \left(x-y\right) \left(x^{2}+xy+y^{2}\right) \)
Therefore, \( \Large \sqrt \frac{ \left(x-y\right) \left(x+y\right) \left(x-y\right) }{ \left(x+y\right)^{3} } = \frac{ \left(x-y\right)^{2} }{ \left(x+y\right)^{2} } \)
= \( \Large \frac{x-y}{x+y} = \frac{0.4-0.2}{0.4+0.02} = \frac{1}{3} \)