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" (iii) "5(1)/(5)+2(1)/(7)...

" (iii) "5(1)/(5)+2(1)/(7)

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Simplify (i) 2^(2/3). 2^(1/3) (ii) (3^(1/5))^4 (iii) (7^(1/5))/(7^(1/3)) (iv) 13^(1/5). 17^(1/5)

Simplify (i) 2^(2/3). 2^(1/3) (ii) (3^(1/5))^4 (iii) (7^(1/5))/(7^(1/3)) (iv) 13^(1/5). 17^(1/5)

(1)/(5) xx [(2)/(7) +(3)/(8)] =[(1)/(5) xx (2)/(7)] + ________.

(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=

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)+…)

If A=[((2)/(3), 1,(5)/(3)),((1)/(3), (2)/(3), (4)/(3)),((7)/(3), 2, (2)/(3))] and B=[((2)/(5), (3)/(5), 1),((1)/(5), (2)/(5), (4)/(5)),((7)/(5),(6)/(5),(2)/(5))] , then compute 3A-5B .

If A=[((2)/(3), 1,(5)/(3)),((1)/(3), (2)/(3), (4)/(3)),((7)/(3), 2, (2)/(3))] and B=[((2)/(5), (3)/(5), 1),((1)/(5), (2)/(5), (4)/(5)),((7)/(5),(6)/(5),(2)/(5))] , then compute 3A-5B .