Math, asked by atharv18259, 5 months ago

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Answered by Chkushu
1

Answer:

\frac{1}{4} ^{-2} -3*8^{\frac{2}{3} } *4^{0} + \frac{9}{16} ^{\frac{-1}{2}}  \\\\4} ^{2}\: -3*\:((8)^{\frac{1}{3} })^{2}*1 \: +\frac{16}{9} ^{\frac{1}{2} } \\\\ 16 - 3*\:(2)^{2} *1\:+ \:\frac{4}{3} \\\\16-3*4*1+\:\frac{4}{3} \\\\16-12+\:\frac{4}{3} \\\\4+\:\frac{4}{3} \\\\\frac{12+4}{3} \\\\\frac{16}{3}

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Hence proved

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Answered by rkcomp31
0

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