Solve - 64125-2-3 - 12566251-4 - -25- 64 0 - 6116

(64125 #160; )^ 23+1 (256625 #160; )^14+ ( √(25)√(64) #160; )^0 = (45 #160; )^3× 23+1 (45 #160; )^4×14+1 = (45 #160; )^ 2+1 (45 ...

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Byju's Answer

Standard IX

Mathematics

Laws of Exponents for Real Numbers

Solve : 6412...

Question

Solve :
${(\frac{64}{125})}^{- 2 / 3} + \frac{1}{{(\frac{256}{625})}^{1 / 4}} + {(\frac{\sqrt{25}}{\sqrt[3]{64}})}^{0} = \frac{61}{16}$

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Solution

${(\frac{64}{125})}^{\frac{- 2}{3}} + \frac{1}{{(\frac{256}{625})}^{\frac{1}{4}}} + {(\frac{\sqrt{25}}{\sqrt[3]{64}})}^{0}$

$= {(\frac{4}{5})}^{3 \times \frac{- 2}{3}} + \frac{1}{{(\frac{4}{5})}^{4 \times \frac{1}{4}}} + 1$

$= {(\frac{4}{5})}^{- 2} + \frac{1}{(\frac{4}{5})} + 1$

$= (\frac{25}{16}) + (\frac{5}{4}) + 1$

$= (\frac{45}{16}) + 1$

$= \frac{61}{16}$

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Similar questions

{(\frac{64}{125})}^{- 2 / 3} \div \frac{1}{{(256 / 625)}^{1 / 4}} + {(\frac{(\sqrt{25})}{\sqrt[3]{64}})}^{0} =

Evaluate: ${(\begin{matrix} \frac{64}{125} \end{matrix})}^{\frac{- 2}{3}} \div \frac{1}{(\frac{256}{625})^{\frac{1}{4}}} + (\begin{matrix} \frac{(\sqrt{25})}{\sqrt[3]{64}} \end{matrix})$

Q. Simplify:
(i)

(27)^{\frac{4}{3}} + (32)^{0.8} + (0.8)^{- 1}

(ii)

{(\frac{64}{125})}^{\frac{- 2}{3}} \div \frac{1}{{(\frac{256}{625})}^{\frac{1}{4}}} \times \frac{\sqrt{25}}{\sqrt[3]{64}}

(iii)

{⎡ ⎢ ⎢ ⎣ 5 {⎛ ⎜ ⎝ 8^{\frac{1}{3}} + 27^{\frac{1}{3}} ⎞ ⎟ ⎠}^{3} ⎤ ⎥ ⎥ ⎦}^{\frac{1}{4}}

Prove that :

$(i) (\sqrt{3 \times 5^{- 3}} \div \sqrt[3]{3^{- 1}} \sqrt{5}) \times \sqrt[6]{3 \times 5^{6}} = \frac{3}{5} (i i) 9^{\frac{3}{2}} - 3 \times 5^{0} - {(\frac{1}{81})}^{\frac{- 1}{2}} = 15 (i i i) {(\frac{1}{4})}^{- 2} - 3 \times 8^{\frac{2}{3}} \times 4^{0} + {(\frac{9}{16})}^{\frac{- 1}{2}} = \frac{16}{3} (i v) \frac{2^{\frac{1}{2}} \times 3^{\frac{1}{3}} \times 4^{\frac{1}{4}}}{10^{\frac{- 1}{5}} \times 5^{\frac{3}{5}}} \div \frac{3^{\frac{4}{3}} \times 5^{\frac{- 7}{5}}}{4^{\frac{- 3}{5}} \times 6} = 10 (v) \sqrt{\frac{1}{4}} + (0.01)^{\frac{- 1}{2}} - (27)^{\frac{2}{3}} = \frac{3}{2} (v i) \frac{2^{n} + 2^{n - 1}}{2^{n + 1} - 2^{n}} = \frac{3}{2} (v i i) {(\frac{64}{125})}^{\frac{- 2}{3}} + \frac{1}{{(\frac{256}{625})}^{\frac{1}{4}}} + {(\frac{\sqrt{25}}{\sqrt[3]{64}})}^{0} = \frac{61}{16} (v i i i) \frac{3^{- 3} \times 6^{2} \times \sqrt{98}}{5^{2} \times \sqrt[3]{\frac{1}{25}} \times (15)^{\frac{- 4}{3}} \times 3^{\frac{1}{3}}} = 28 \sqrt{2} (i x) \frac{(0.6)^{0} - (0.1)^{- 1}}{{(\frac{3}{8})}^{- 1} {(\frac{3}{2})}^{3} + {(\frac{1}{3})}^{- 1}} = - \frac{3}{2}$

Q. 3, 10, 27, 4, 16, 64, 5, 25, 125

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Solve :(64125)−2/3+1(256625)1/4+(√253√64)0=6116

(64125)−23+1(256625)14+(√253√64)0=(45)3×−23+1(45)4×14+1=(45)−2+1(45)+1=(2516)+(54)+1=(4516)+1=6116

Solve :
${(\frac{64}{125})}^{- 2 / 3} + \frac{1}{{(\frac{256}{625})}^{1 / 4}} + {(\frac{\sqrt{25}}{\sqrt[3]{64}})}^{0} = \frac{61}{16}$