Question

Simplify
(i)(2)^2/3×(2)^1/5
(ii)(7)^1/5
(iii)(32)^2/5
(iv)(125)^1/5​

Answer 1

Answer:

Step-by-step explanation:

Hi ,

write 81 ,216 , 32 ,225 as product of the prime.

1 )  81 = 3 × 3 × 3 × 3  

2 ) 216 = 2 × 2 × 2 × 3 × 3 × 3

3 )   32 = 2 × 2 × 2 × 2 × 2

4 ) 225 = 3 × 3 × 5 × 5

Now ,

i ) fourth root of 81 = ( 81 ) 1/4 = ( 3^4 ) 1/4 = 3 ^ (4 × 1/4 ) = 3

ii) ∛216 = ( 216 ) 1/3 = ( 2³ × 3³ )^ 1/3 = [ ( 2 × 3 )³ ]^ 1/3 = (2 × 3 ) = 6

iii ) fifth root 0f 32 = ( 32 )^1/5 = ( 2^5 ) ^1/5 = 2

iv ) √225 = √ ( 3 × 3 ) × ( 5 × 5 ) = 3 × 5 = 15

Given problem is ,

fourth root of 81 - 8 ∛216 + 15 fifth root of 32 + √225

= 3 - ( 8 × 6 ) +  ( 15 × 2 ) + 15  [ put from ( i ) to ( iv ) values ]

= 3 - 48 + 30 +15

= 48 - 48

= 0

hope it helps

:)

Answer 2

Answer:

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Step-by-step explanation:

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