[17]−3×7−1×[149]\Big[\dfrac{1}{7}\Big]^{-3} \times 7^{-1} \times \Big[\dfrac{1}{49}\Big][71]−3×7−1×[491] is equal to:
-1
17\dfrac{1}{7}71
-7
1
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[17]−3×7−1×[149]=[71]3×[17]1×[149]=[7×7×71×1×1]×[17]×[149]=[3431]×[17]×[149]=[343×1×11×7×49]=[343343]=1\Big[\dfrac{1}{7}\Big]^{-3} \times 7^{-1} \times \Big[\dfrac{1}{49}\Big]\\[1em] = \Big[\dfrac{7}{1}\Big]^3 \times \Big[\dfrac{1}{7}\Big]^1 \times \Big[\dfrac{1}{49}\Big]\\[1em] = \Big[\dfrac{7 \times 7 \times 7}{1 \times 1 \times 1}\Big] \times \Big[\dfrac{1}{7}\Big] \times \Big[\dfrac{1}{49}\Big]\\[1em] = \Big[\dfrac{343}{1}\Big] \times \Big[\dfrac{1}{7}\Big] \times \Big[\dfrac{1}{49}\Big]\\[1em] = \Big[\dfrac{343 \times 1 \times 1}{1 \times 7 \times 49}\Big]\\[1em] = \Big[\dfrac{343}{343}\Big]\\[1em] = 1[71]−3×7−1×[491]=[17]3×[71]1×[491]=[1×1×17×7×7]×[71]×[491]=[1343]×[71]×[491]=[1×7×49343×1×1]=[343343]=1
Hence, option 4 is the correct option.
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If [45]−3×[45]−5=[45]3x−2\Big[\dfrac{4}{5}\Big]^{-3} \times \Big[\dfrac{4}{5}\Big]^{-5} = \Big[\dfrac{4}{5}\Big]^{3x - 2}[54]−3×[54]−5=[54]3x−2, the value of x is:
2
12\dfrac{1}{2}21
-2
−12-\dfrac{1}{2}−21
If [mn]x−1=[nm]x−5\Big[\dfrac{m}{n}\Big]^{x-1} = \Big[\dfrac{n}{m}\Big]^{x-5}[nm]x−1=[mn]x−5, the value of x is:
3
-3
13\dfrac{1}{3}31
−13-\dfrac{1}{3}−31
Compute:
18×30×53×221^8 \times 3^0 \times 5^3 \times 2^218×30×53×22
(47)2×(4−3)4(4^7)^2 \times (4^{-3})^4(47)2×(4−3)4
