64 by 125 to the power minus 2 by 3 + 256 by 625 to the power minus 1 by 4 + 3 by 7 to the power 0
Answers
Answered by
4
Hi ,
(64/125)^-2/3+(256/625)^1/4 +(3/7)^0
= (125/64)^2/3+ ( 256/625)^1/4 + 1
[ Since 1 ) ( a/b )^-n = ( b/a )ⁿ
2 ) a^0 = 1 ]
= ( 5³/4³ )^2/3 + ( 4⁴/5⁴ )1/4 + 1
= [ ( 5/4 )³ ]^2/3 + [ ( 4/5 )⁴ ]^1/4 + 1
= ( 5/4 ) ^3×2/3 + ( 4/5 )^4×1/4 + 1
= ( 5/4 )² + ( 4/5 ) + 1
= 25/16 + 4/5 + 1
= ( 125 + 64 + 80 ) / 80
= 269/80
I hope this helps you.
: )
(64/125)^-2/3+(256/625)^1/4 +(3/7)^0
= (125/64)^2/3+ ( 256/625)^1/4 + 1
[ Since 1 ) ( a/b )^-n = ( b/a )ⁿ
2 ) a^0 = 1 ]
= ( 5³/4³ )^2/3 + ( 4⁴/5⁴ )1/4 + 1
= [ ( 5/4 )³ ]^2/3 + [ ( 4/5 )⁴ ]^1/4 + 1
= ( 5/4 ) ^3×2/3 + ( 4/5 )^4×1/4 + 1
= ( 5/4 )² + ( 4/5 ) + 1
= 25/16 + 4/5 + 1
= ( 125 + 64 + 80 ) / 80
= 269/80
I hope this helps you.
: )
Answered by
0
Answer:
(64/125)^-2/3 + (256/625)^-1/4 + (3/7)^0
= (4^3/5^3)^-2/3 + (4^4/5^4)^-1/4 + 1
= (4/5)^-2 + (4/5)^-1 + 1
= (5/4)^2 + (5/4)^1 + 1
= 25/16 + 5/4 + 1
= 25 + 20 + 16 / 16
= 61 / 16 (Ans)
Similar questions