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Mathematics

Evaluate:

90+9−1−9−2+912−9−129^0 + 9^{-1} - 9^{-2} + 9^{\dfrac{1}{2}} - 9^{-\dfrac{1}{2}}

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90+9−1−9−2+912−9−12=1+191−192+(32)12−1912=1+191−192+(3)2×12−123212=1+191−192+(3)1−132×12=1+19−192+(3)−131=1+19−19×9+(3)−13=4+19−181−13=41+19−181−139^0 + 9^{-1} - 9^{-2} + 9^{\dfrac{1}{2}} - 9^{-\dfrac{1}{2}}\\[1em] = 1 + \dfrac{1}{9}^1 - \dfrac{1}{9}^2 + (3^2)^{\dfrac{1}{2}} -\dfrac{1}{9}^{\dfrac{1}{2}}\\[1em] = 1 + \dfrac{1}{9}^1 - \dfrac{1}{9}^2 + (3)^{2\times\dfrac{1}{2}} -\dfrac{1^2}{3^2}^{\dfrac{1}{2}}\\[1em] = 1 + \dfrac{1}{9}^1 - \dfrac{1}{9}^2 + (3)^1 -\dfrac{1}{3}^{2\times\dfrac{1}{2}}\\[1em] = 1 + \dfrac{1}{9} - \dfrac{1}{9}^2 + (3) -\dfrac{1}{3}^1\\[1em] = 1 + \dfrac{1}{9} - \dfrac{1}{9\times 9} + (3) -\dfrac{1}{3}\\[1em] = 4 + \dfrac{1}{9} - \dfrac{1}{81} -\dfrac{1}{3}\\[1em] = \dfrac{4}{1} + \dfrac{1}{9} - \dfrac{1}{81} -\dfrac{1}{3}

LCM of 1, 9, 81 and 3 is 3 x 3 x 3 x 3 =81

=4×811×81+1×99×9−1×181×1−1×273×27=32481+981−181−2781=324+9−1−2781=30581=36281= \dfrac{4 \times 81}{1 \times 81} + \dfrac{1 \times 9}{9 \times 9} - \dfrac{1 \times 1}{81 \times 1} -\dfrac{1 \times 27}{3 \times 27}\\[1em] = \dfrac{324}{81} + \dfrac{9}{81} - \dfrac{1}{81} -\dfrac{27}{81}\\[1em] = \dfrac{324 + 9 -1 -27}{81}\\[1em] = \dfrac{305}{81}\\[1em] = 3\dfrac{62}{81}

90+9−1−9−2+912−9−12=362819^0 + 9^{-1} - 9^{-2} + 9^{\dfrac{1}{2}} - 9^{-\dfrac{1}{2}} = 3\dfrac{62}{81}

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