How do you evaluate ${[\frac{3}{2} - {(\frac{1}{2} - \frac{3}{4})}^{2}]}^{- 1} + {[{(3 + \frac{1}{3})}^{2} - {(3 - \frac{1}{3})}^{2}]}^{- \frac{1}{2}}$ $[\frac { 3} { 2} - ( \frac { 1} { 2} - \frac { 3} { 4} ) ^ { 2} ] ^ { - 1} + [ ( 3+ \frac { 1} { 3} ) ^ { 2} - ( 3- \frac { 1} { 3} ) ^ { 2} ] ^ { - \frac { 1} { 2} }$ ?

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How do you evaluate ${[\frac{3}{2} - {(\frac{1}{2} - \frac{3}{4})}^{2}]}^{- 1} + {[{(3 + \frac{1}{3})}^{2} - {(3 - \frac{1}{3})}^{2}]}^{- \frac{1}{2}}$ $[\frac { 3} { 2} - ( \frac { 1} { 2} - \frac { 3} { 4} ) ^ { 2} ] ^ { - 1} + [ ( 3+ \frac { 1} { 3} ) ^ { 2} - ( 3- \frac { 1} { 3} ) ^ { 2} ] ^ { - \frac { 1} { 2} }$ ?

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Answer 1 · 2018-03-29T15:39:04

${[\frac{3}{2} - {(\frac{2 - 3}{4})}^{2}]}^{- 1} + {[{(\frac{9 + 1}{3})}^{2} - {(\frac{9 - 1}{3})}^{2}]}^{- \frac{1}{2}}$ $[3/2-((2-3)/4)^2]^-1+[((9+1)/3)^2-((9-1)/3)^2]^(-1/2)$

Explanation:

$= {[\frac{3}{2} - {(\frac{- 1}{4})}^{2}]}^{- 1} + {[{(\frac{10}{3})}^{2} - {(\frac{8}{3})}^{2}]}^{- \frac{1}{2}} =$ $=[3/2-((-1)/4)^2]^-1+[((10)/3)^2-((8)/3)^2]^(-1/2)=$
$= {[\frac{3}{2} - \frac{1}{8}]}^{- 1} + {[\frac{100}{9} - \frac{64}{9}]}^{- \frac{1}{2}} =$ $=[3/2-1/8]^-1+[100/9-64/9]^(-1/2)=$
$= {[\frac{12 - 1}{8}]}^{- 1} + {[\frac{100 - 64}{9}]}^{- \frac{1}{2}} =$ $=[(12-1)/8]^-1+[(100-64)/9]^(-1/2)=$
$= {[\frac{11}{8}]}^{- 1} + {[\frac{36}{9}]}^{- \frac{1}{2}} =$ $=[11/8]^-1+[36/9]^(-1/2)=$
$= \frac{1}{\frac{8}{11}} + \frac{1}{{[\frac{36}{9}]}^{\frac{1}{2}}} =$ $=1/[8/11]+1/[36/9]^(1/2)=$
$= \frac{11}{8} + {[\frac{9}{36}]}^{\frac{1}{2}} =$ $=11/8+[9/36]^(1/2)=$
$= \frac{11}{8} + {[\frac{3^{2}}{6^{2}}]}^{\frac{1}{2}} =$ $=11/8+[3^2/6^2]^(1/2)=$
$= \frac{11}{8} + \frac{3}{6} = \frac{11}{8} + \frac{1}{2} = \frac{11 + 4}{8} = \frac{15}{8}$ $=11/8+3/6=11/8+1/2=(11+4)/8=15/8$

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