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Evaluate : (64)^(2/3) - Root(3)(125) - 1/2^(-5) + (27)^(-2/3) Xx (25/9)^(-1/2) - Mathematics

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प्रश्न

Evaluate : `(64)^(2/3) - root(3)(125) - 1/2^(-5) + (27)^(-2/3) xx (25/9)^(-1/2)`

योग

उत्तर

`(64)^(2/3) - root(3)(125) - 1/2^(-5) + (27)^(-2/3) xx (25/9)^(-1/2)`

= `(4^3)^(2/3) - root(3)(5^3) - 2^5 + (3^3)^(-2/3) xx ((5^2)/(3^2))^(-1/2)`

= `4^2 - 5 - 2^5 + 3^-2 xx (5/3)^( 2 xx (-1/2)`

= `16 - 5 - 32 + 1/3^2 xx (5/3)^-1`

= `- 21 + 1/9 xx 3/5`

= `- 21 + 1/15`

= `[ - 315 + 1 ]/15`

= `- 314/15`

= `- 20 14/15`

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अध्याय 7: Indices (Exponents) - Exercise 7 (C) [पृष्ठ १०१]

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सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 7 Indices (Exponents)
Exercise 7 (C) | Q 1.2 | पृष्ठ १०१
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