Simplify the following and express with positive index :
(27−39−3)15\Big(\dfrac{27^{-3}}{9^{-3}}\Big)^{\dfrac{1}{5}}(9−327−3)51
27 Likes
Simplifying the expression :
⇒(27−39−3)15=[(33)−3(32)−3]15=(33×−332×−3)15=(3−93−6)15=(3−9−(−6))15=(3−9+6)15=(3−3)15=(3)−35=(133)15=1335.\Rightarrow \Big(\dfrac{27^{-3}}{9^{-3}}\Big)^{\dfrac{1}{5}} = \Big[\dfrac{(3^3)^{-3}}{(3^2)^{-3}}\Big]^{\dfrac{1}{5}}\\[1em] = \Big(\dfrac{3^{3 \times -3}}{3^{2 \times -3}}\Big)^{\dfrac{1}{5}} = \Big(\dfrac{3^{-9}}{3^{-6}}\Big)^{\dfrac{1}{5}} \\[1em] = (3^{-9 - (-6)})^{\dfrac{1}{5}} = (3^{-9 + 6})^{\dfrac{1}{5}} \\[1em] = (3^{-3})^{\dfrac{1}{5}} = (3)^{-\dfrac{3}{5}} \\[1em] = \Big(\dfrac{1}{3^3}\Big)^{\dfrac{1}{5}} = \dfrac{1}{3^{\dfrac{3}{5}}}.⇒(9−327−3)51=[(32)−3(33)−3]51=(32×−333×−3)51=(3−63−9)51=(3−9−(−6))51=(3−9+6)51=(3−3)51=(3)−53=(331)51=3531.
Hence, (27−39−3)15=1335\Big(\dfrac{27^{-3}}{9^{-3}}\Big)^{\dfrac{1}{5}} = \dfrac{1}{3^{\dfrac{3}{5}}}(9−327−3)51=3531.
Answered By
18 Likes
Evaluate :
(278)23−(14)−2+50\Big(\dfrac{27}{8}\Big)^{\dfrac{2}{3}} - \Big(\dfrac{1}{4}\Big)^{-2} + 5^0(827)32−(41)−2+50
(3−42−8)14\Big(\dfrac{3^{-4}}{2^{-8}}\Big)^{\dfrac{1}{4}}(2−83−4)41
(32)−25÷(125)−23(32)^{-\dfrac{2}{5}} ÷ (125)^{-\dfrac{2}{3}}(32)−52÷(125)−32
[1 - {1 - (1 - n)-1}-1]-1
