Evaluate:
(2−3+3−2)×70(2^{-3} + 3^{-2})\times 7^0(2−3+3−2)×70
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(2−3+3−2)×70=(123+132)×1=(18+19)(2^{-3} + 3^{-2})\times 7^0\\[1em] = \Big(\dfrac{1}{2^{3}} + \dfrac{1}{3^{2}}\Big)\times 1\\[1em] = \Big(\dfrac{1}{8} + \dfrac{1}{9}\Big)(2−3+3−2)×70=(231+321)×1=(81+91)
LCM of 8 and 9 is 2 x 2 x 2 x 3 x 3 = 72
=(1×98×9+1×89×8)=(972+872)=(9+872)=(1772)= \Big(\dfrac{1 \times 9}{8 \times 9} + \dfrac{1 \times 8}{9 \times 8}\Big)\\[1em] = \Big(\dfrac{9}{72} + \dfrac{8}{72}\Big)\\[1em] = \Big(\dfrac{9 + 8}{72}\Big)\\[1em] = \Big(\dfrac{17}{72}\Big)=(8×91×9+9×81×8)=(729+728)=(729+8)=(7217)
(2−3+3−2)×70=(1772)(2^{-3} + 3^{-2})\times 7^0 = \Big(\dfrac{17}{72}\Big)(2−3+3−2)×70=(7217)
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Simplify:
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x2n+7.(x2)3n+2x4(2n+3)\dfrac{x^{2n+7}.(x^2)^{3n+2}}{x^{4(2n+3)}}x4(2n+3)x2n+7.(x2)3n+2
(80+2−1)×32(8^0 + 2^{-1})\times 3^2(80+2−1)×32
[(16)−1−(15)−1]−2\Big[\Big(\dfrac{1}{6}\Big)^{-1} - \Big(\dfrac{1}{5}\Big)^{-1}\Big]^{-2}[(61)−1−(51)−1]−2
