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Simplify: (i) 22/3×21/5,(ii) (1/33)7,(iii) 111/2/111/4,(iv) 71/2×81 .Q.3(i),(ii),(iii),(iv)

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Today i am solving the ncert class 9th chapter number system . How i solve the exercise 1.6 question number 3 . Give me the best and easiest solution of this question also find the best method of this question.Simplify: (i) 22/3×21/5,(ii) (1/33)7,(iii) 111/2/111/4,(iv) 71/2×81 .

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  1. (i) 22/3×21/5

    Solution:

    22/3×21/5 = 2(2/3)+(1/5) [⸪Since, am×an=am+n____ Laws of exponents]

    = 213/15 [⸪2/3 + 1/5 = (2×5+3×1)/(3×5) = 13/15]

    (ii) (1/33)7

    Solution:

    (1/33)= (3-3)7 [⸪Since,(am)= am x n____ Laws of exponents]

    = 3-21

    (iii) 111/2/111/4

    Solution:

    111/2/111/4 = 11(1/2)-(1/4)

    = 111/4 [⸪(1/2) – (1/4) = (1×4-2×1)/(2×4) = 4-2)/8 = 2/8 = ¼ ]

    (iv) 71/2×81/2

    Solution:

    71/2×81/2 = (7×8)1/2 [⸪Since, (am×b= (a×b)m ____ Laws of exponents]

    = 561/2

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