" (iii) "(7^((1)/(5)))/(7^((1)/(3)))

Simplify: ( 7^((1)/(5)))/( 7^((1)/(3)))

Simplify 7^(1/5)/7^(1/3)

The value of: [(5/7 " of " 1 (3)/(8) " of " 6/7)] div [1-(1)/(7) xx ( 5/12 +(1)/(3)) ] xx ((1)/(7) -(1)/(9))/((1)/(7) +(1)/(9))

(1)/(2)((1)/(5)+(1)/(7))-(1)/(4)((1)/(5^(2))+(1)/(7^(2)))+(1)/(6)((1)/(5^(3))+(1)/(7^(3)))-….oo=

(1)/(2)((1)/(5)+(1)/(7))-(1)/(4)((1)/(5^(2))+(1)/(7^(2)))+(1)/(6)((1)/(5^(3))+(1)/(7^(3)))-….oo=

The greatest of the numbers 2(1)/(2),3^((1)/(3)),4^((1)/(4)),5^((1)/(5)),6^((1)/(6)) and 7^((1)/(7)) is

The number (7+5sqrt(2))^((1)/(3))+(7-5sqrt(2))^((1)/(3)) ,is equal to

Show that A= [(5,3),(-1,-2)] satisfies the equation x^2-3x-7=0 . Thus, find A^(-1) . a) [((2)/(7),(3)/(7)),((-1)/(7),(-5)/(7))] b) [(2,3),(-1,-5)] c) [((1)/(7),(1)/(7)),((-1)/(7),(-5)/(7))] d) [(3,-1),(1,2)]

Assertion (A) : (1)/(5)+(1)/(3.5^(3))+(1)/(5.5^(5))+(1)/(7.5^(7))+…(1)/(2)log((3)/(2)) Reason (R ) : If |x| lt 1 then log_(e )((1+x)/(1-x))=2(x+(x^(3))/(3)+(x^(5))/(5)+…)