$\begin{aligned} [8 - {(\frac{4^{\frac{9}{4}} \sqrt{{2.2}^{2}}}{2 \sqrt{2^{- 2}}})}^{^{\frac{1}{2}}}] \\ is equal to = ? \end{aligned}$ $\eqalign{ & \left[ {8 - {{\left( {\frac{{{4^{\frac{9}{4}}}\sqrt {{{2.2}^2}} }}{{2\sqrt {{2^{ - 2}}} }}} \right)}^{^{\frac{1}{2}}}}} \right] \cr & {\text{is equal to = ?}} \cr}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} [8 - {(\frac{4^{\frac{9}{4}} \sqrt{{2.2}^{2}}}{2 \sqrt{2^{- 2}}})}^{^{\frac{1}{2}}}] \\ = [8 - {(\frac{2^{2 \times \frac{9}{4}} \sqrt{2^{1 + 2}}}{2 \sqrt{\frac{1}{4}}})}^{^{\frac{1}{2}}}] \\ = [8 - {(\frac{2^{\frac{9}{2}} {.2}^{\frac{3}{2}}}{2 \times \frac{1}{2}})}^{^{\frac{1}{2}}}] \\ = [8 - {(2^{\frac{12}{2}})}^{^{\frac{1}{2}}}] \\ = [8 - (2^{6 \times \frac{1}{2}})] \\ = [8 - (2^{3})] \\ = [8 - 8] \\ = 0 \end{aligned}

$\eqalign{ & \left[ {8 - {{\left( {\frac{{{4^{\frac{9}{4}}}\sqrt {{{2.2}^2}} }}{{2\sqrt {{2^{ - 2}}} }}} \right)}^{^{\frac{1}{2}}}}} \right] \cr & = \left[ {8 - {{\left( {\frac{{{2^{2 \times \frac{9}{4}}}\sqrt {{2^{1 + 2}}} }}{{2\sqrt {\frac{1}{4}} }}} \right)}^{^{\frac{1}{2}}}}} \right] \cr & = \left[ {8 - {{\left( {\frac{{{2^{\frac{9}{2}}}{{.2}^{\frac{3}{2}}}}}{{2 \times \frac{1}{2}}}} \right)}^{^{\frac{1}{2}}}}} \right] \cr & = \left[ {8 - {{\left( {{2^{\frac{{12}}{2}}}} \right)}^{^{\frac{1}{2}}}}} \right] \cr & = \left[ {8 - \left( {{2^{6 \times \frac{1}{2}}}} \right)} \right] \cr & = \left[ {8 - \left( {{2^3}} \right)} \right] \cr & = \left[ {8 - 8} \right] \cr & = 0 \cr}$