Compute:
(−3)4−(34)0×(−2)5÷(64)23(-3)^4 - (\sqrt[4]{3})^0 \times (-2)^5 ÷ (64)^{\dfrac{2}{3}}(−3)4−(43)0×(−2)5÷(64)32
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As we know for any rational number,
a0=1a^0 = 1a0=1
(−3)4−(34)0×(−2)5÷(64)23=(−3×−3×−3×−3)−1×(−2)5÷(43)23=81−(−2×−2×−2×−2×−2)÷(4)3×23=81−(−32)÷(4)2=81+32÷(4×4)=81+32÷16=81+2=83(-3)^4 - (\sqrt[4]{3})^0 \times (-2)^5 ÷ (64)^{\dfrac{2}{3}}\\[1em] =(-3 \times -3 \times -3 \times -3) - 1 \times (-2)^5 ÷ (4^3)^{\dfrac{2}{3}}\\[1em] = 81 - (-2 \times -2 \times -2 \times -2 \times -2) ÷ (4)^{3\times\dfrac{2}{3}}\\[1em] = 81 - (-32) ÷ (4)^{2}\\[1em] = 81 + 32 ÷ (4 \times 4)\\[1em] = 81 + 32 ÷ 16\\[1em] = 81 + 2\\[1em] = 83(−3)4−(43)0×(−2)5÷(64)32=(−3×−3×−3×−3)−1×(−2)5÷(43)32=81−(−2×−2×−2×−2×−2)÷(4)3×32=81−(−32)÷(4)2=81+32÷(4×4)=81+32÷16=81+2=83
(−3)4−(34)0×(−2)5÷(64)23=83(-3)^4 - (\sqrt[4]{3})^0 \times (-2)^5 ÷ (64)^{\dfrac{2}{3}} = 83(−3)4−(43)0×(−2)5÷(64)32=83
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(125)−23÷(8)23(125)^{-\dfrac{2}{3}} ÷ (8)^{\dfrac{2}{3}}(125)−32÷(8)32
(243)25÷(32)−25(243)^{\dfrac{2}{5}} ÷ (32)^{-\dfrac{2}{5}}(243)52÷(32)−52
(27)23÷(8116)−14(27)^{\dfrac{2}{3}} ÷ \Big(\dfrac{81}{16}\Big)^{-\dfrac{1}{4}}(27)32÷(1681)−41
Simplify:
843+2532−(127)−238^{\dfrac{4}{3}} + 25^{\dfrac{3}{2}} - \Big(\dfrac{1}{27}\Big)^{-\dfrac{2}{3}}834+2523−(271)−32
