(−5)5×(−5)−3(-5)^5 \times (-5)^{-3}(−5)5×(−5)−3 is equal to:
15\dfrac{1}{5}51
5
-25
25
4 Likes
As we know, for any non-zero rational number a
a−n=1ana^{-n} = \dfrac{1}{a^n}a−n=an1 and an=1a−na^{n} = \dfrac{1}{a^{-n}}an=a−n1.
(−5)5×(−5)−3=(−5)5×(−15)3=(−5×−5×−5×−5×−5)×(−1×−1×−15×5×5)=−3125×(−1125)=(−3125×−1125)=(3125125)=25(-5)^5 \times (-5)^{-3}\\[1em] = (-5)^5 \times \Big(\dfrac{-1}{5}\Big)^3\\[1em] = (-5 \times -5 \times -5 \times -5 \times -5) \times \Big(\dfrac{-1 \times -1 \times -1}{5 \times 5 \times 5}\Big)\\[1em] = -3125 \times \Big(\dfrac{-1}{125}\Big)\\[1em] = \Big(\dfrac{-3125 \times -1}{125}\Big)\\[1em] = \Big(\dfrac{3125}{125}\Big)\\[1em] = 25(−5)5×(−5)−3=(−5)5×(5−1)3=(−5×−5×−5×−5×−5)×(5×5×5−1×−1×−1)=−3125×(125−1)=(125−3125×−1)=(1253125)=25
Hence, option 4 is the correct option.
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(23)3×(32)6\Big(\dfrac{2}{3}\Big)^3 \times \Big(\dfrac{3}{2}\Big)^6(32)3×(23)6 is equal to:
827\dfrac{8}{27}278
278\dfrac{27}{8}827
49\dfrac{4}{9}94
94\dfrac{9}{4}49
80+8−1+4−18^0 + 8^{-1} + 4^{-1}80+8−1+4−1 is equal to:
8388\dfrac{3}{8}883
1381\dfrac{3}{8}183
38\dfrac{3}{8}83
2232\dfrac{2}{3}232
Evaluate:
(3−1×9−1)÷3−2(3^{-1} \times 9^{-1}) ÷ 3^{-2}(3−1×9−1)÷3−2
(3−1×4−1)÷6−1(3^{-1} \times 4^{-1}) ÷ 6^{-1}(3−1×4−1)÷6−1
