Evaluate:
(14)−2−3(32)25×(7)0+(916)−12\Big(\dfrac{1}{4}\Big)^{-2} - 3(32)^\dfrac{2}{5}\times (7)^0 + \Big(\dfrac{9}{16}\Big)^{-\dfrac{1}{2}}(41)−2−3(32)52×(7)0+(169)−21
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(14)−2−3(32)25×(7)0+(916)−12=(41)2−3(25)25×1+(169)12=16−3(2)2×55+(4232)12=16−3(2)2+(43)1×22=16−3×4+43=16−12+43=4+43=12+43=163=513\Big(\dfrac{1}{4}\Big)^{-2} - 3(32)^\dfrac{2}{5}\times (7)^0 + \Big(\dfrac{9}{16}\Big)^{-\dfrac{1}{2}}\\[1em] = \Big(\dfrac{4}{1}\Big)^{2} - 3(2^5)^\dfrac{2}{5}\times 1 + \Big(\dfrac{16}{9}\Big)^{\dfrac{1}{2}}\\[1em] = 16 - 3(2)^\dfrac{2 \times 5}{5} + \Big(\dfrac{4^2}{3^2}\Big)^{\dfrac{1}{2}}\\[1em] = 16 - 3(2)^2 + \Big(\dfrac{4}{3}\Big)^{\dfrac{1 \times 2}{2}}\\[1em] = 16 - 3 \times 4 + \dfrac{4}{3}\\[1em] = 16 - 12 + \dfrac{4}{3}\\[1em] = 4 + \dfrac{4}{3}\\[1em] = \dfrac{12 + 4}{3}\\[1em] = \dfrac{16}{3}\\[1em] = 5\dfrac{1}{3}(41)−2−3(32)52×(7)0+(169)−21=(14)2−3(25)52×1+(916)21=16−3(2)52×5+(3242)21=16−3(2)2+(34)21×2=16−3×4+34=16−12+34=4+34=312+4=316=531
Hence, (14)−2−3(32)25×(7)0+(916)−12=513\Big(\dfrac{1}{4}\Big)^{-2} - 3(32)^\dfrac{2}{5}\times (7)^0 + \Big(\dfrac{9}{16}\Big)^{-\dfrac{1}{2}} = 5\dfrac{1}{3}(41)−2−3(32)52×(7)0+(169)−21=531.
Answered By
Given (827)x−1\Big(\dfrac{8}{27}\Big)^{x-1}(278)x−1 = (94)2x+1\Big(\dfrac{9}{4}\Big)^{2x+1}(49)2x+1; find the value of x .
14+(0.01)−12×(5)−(27)23\sqrt\dfrac{1}{4} + (0.01)^{-\dfrac{1}{2}} \times (5) - (27)^\dfrac{2}{3}41+(0.01)−21×(5)−(27)32
If xa=yb=zcx^a = y^b = z^cxa=yb=zc and y2=xzy^2 = xzy2=xz; prove that b=2aca+cb = \dfrac{2ac}{a+c}b=a+c2ac.
1(216)−23÷1(27)−43\dfrac{1}{(216)^{-\dfrac{2}{3}}} \div \dfrac{1}{(27)^{-\dfrac{4}{3}}}(216)−321÷(27)−341
