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Evaluate : (27/8)^(2/3) - (1/4)^-2 + 5^0 - Mathematics

Sum

Evaluate :
`(27/8)^(2/3) - (1/4)^-2 + 5^0`

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Solution

`( 27/8 )^(2/3) - (1/4)^(-2) + 5^0`

= `([ 3 xx 3 xx 3]/[ 2 xx 2 xx 2 ])^(2/3) - ([ 1 xx 1 ]/[ 2 xx 2 ])^-2 + 5^0`

= `[(3/2)^3]^(2/3) - [(1/2)^2]^-2 + 1`

= `(3/2)^( 3 xx 2/3 ) - (1/2)^[2 xx ( - 2)] + 1`

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

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

= `9/4 - 1/[ 1/2 xx 1/2 xx 1/2 xx 1/2 ] + 1`

= `9/4 - 1/[1/16] + 1`

= `9/4 - 16 + 1`

= `[ 9 - 64 + 4 ]/4`

= `(-51)/4`

Concept: Laws of Exponents
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APPEARS IN

Selina Concise Mathematics Class 9 ICSE
Chapter 7 Indices (Exponents)
Exercise 7 (A) | Q 3.2 | Page 98
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