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Question
secondary 4 | A Maths
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please help thankyouu
Date Posted: 4 years ago
Views: 207
³√a / a-¹ ÷ √a³
= a^⅓ × a¹ ÷ (a³)^½
= a^⅓ × a¹ ÷ a^³⁄²
= a^(⅓ + 1 - 3/2)
= a^(-1/6)
= 1/( a^(1/6) )
= 1/6√a
(Note the 6√ here is written on paper like ³√. It's a superscript. My keyboard doesn't have superscript for 6 so couldn't type it like ³
= a^⅓ × a¹ ÷ (a³)^½
= a^⅓ × a¹ ÷ a^³⁄²
= a^(⅓ + 1 - 3/2)
= a^(-1/6)
= 1/( a^(1/6) )
= 1/6√a
(Note the 6√ here is written on paper like ³√. It's a superscript. My keyboard doesn't have superscript for 6 so couldn't type it like ³
See 1 Answer
done
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Note that the question is worth 6 marks, so its very likely that you cannot leave a surd term in the denominator, especially for Amaths.
Date Posted:
4 years ago
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