Evaluate:
1(216)−23÷1(27)−43\dfrac{1}{(216)^{-\dfrac{2}{3}}} \div \dfrac{1}{(27)^{-\dfrac{4}{3}}}(216)−321÷(27)−341
1(216)−23÷1(27)−43=21623÷2743=(63)23÷(33)43=(6)2×33÷(3)4×33=62÷34=3681=49\dfrac{1}{(216)^{-\dfrac{2}{3}}} \div \dfrac{1}{(27)^{-\dfrac{4}{3}}}\\[1em] = 216^{\dfrac{2}{3}} \div 27^{\dfrac{4}{3}}\\[1em] = (6^3)^{\dfrac{2}{3}} \div (3^3)^{\dfrac{4}{3}}\\[1em] = (6)^{\dfrac{2 \times 3}{3}} \div (3)^{\dfrac{4 \times 3}{3}}\\[1em] = 6^2 \div 3^4\\[1em] = \dfrac{36}{81}\\[1em] = \dfrac{4}{9}(216)−321÷(27)−341=21632÷2734=(63)32÷(33)34=(6)32×3÷(3)34×3=62÷34=8136=94
Hence, 1(216)−23÷1(27)−43=49\dfrac{1}{(216)^{-\dfrac{2}{3}}} \div \dfrac{1}{(27)^{-\dfrac{4}{3}}} = \dfrac{4}{9}(216)−321÷(27)−341=94.
Answered By
(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
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.
[5(813+2713)3]14\Big[5\Big(8^{\dfrac{1}{3}}+ 27^{\dfrac{1}{3}}\Big)^3\Big]^{\dfrac{1}{4}}[5(831+2731)3]41
If (314)4−(413)4(314)2−(413)2=(13a12)2\dfrac{\Big(3\dfrac{1}{4}\Big)^4 - \Big(4\dfrac{1}{3}\Big)^4}{\Big(3\dfrac{1}{4}\Big)^2 - \Big(4\dfrac{1}{3}\Big)^2} = \Big(\dfrac{13a}{12}\Big)^2(341)2−(431)2(341)4−(431)4=(1213a)2, find a.
