Bihar Board Class 9 Maths Solutions Chapter 1 संख्या पद्धति Ex 1.6

Bihar Board Class 9 Maths Solutions Chapter 1 संख्या पद्धति Ex 1.6 Text Book Questions and Answers.

BSEB Bihar Board Class 9 Maths Solutions Chapter 1 संख्या पद्धति Ex 1.6

प्रश्न 1.
ज्ञात कीजिए
(i) 64\(\frac{1}{2}\)
(ii) 32\(\frac{1}{5}\)
(iii) 125\(\frac{1}{3}\)
हाल:
(i) (64)\(\frac{1}{2}\) = (8 × 8)\(\frac{1}{2}\)
= (8²)\(\frac{1}{2}\) [∵ (xa)b = xab]
= 8

Bihar Board Class 9 Maths Solutions Chapter 1 संख्या पद्धति Ex 1.6

(ii) 32\(\frac{1}{5}\) = (2 × 2 × 2 × 2 × 2)\(\frac{1}{5}\)
= (25)\(\frac{1}{5}\)
= 25×\(\frac{1}{5}\) [∵ (xa)b = xab]
= 2

(iii) 125\(\frac{1}{3}\) = (5 × 5 × 5)\(\frac{1}{3}\)
= (53)\(\frac{1}{3}\)
= 53×\(\frac{1}{3}\) [∵ (xa)b = xab]
= 5

Bihar Board Class 9 Maths Solutions Chapter 1 संख्या पद्धति Ex 1.6

प्रश्न 2.
ज्ञात कीजिए
(i) 9\(\frac{3}{2}\)
(ii) 32\(\frac{2}{5}\)
(iii) 16\(\frac{3}{4}\)
(iv) 125\(\frac{-1}{3}\)
हल:
(i) 9\(\frac{3}{2}\) = (3 × 3)\(\frac{3}{2}\)
= (32)\(\frac{3}{2}\)
= 32×\(\frac{3}{2}\) [∵ (xa)b = xab]
= 3³
= 27

Bihar Board Class 9 Maths Solutions Chapter 1 संख्या पद्धति Ex 1.6

(ii) 32\(\frac{2}{5}\) = (2 × 2 × 2 × 2 × 2)\(\frac{2}{5}\)
= (25)\(\frac{2}{5}\)
= 25×\(\frac{2}{5}\) [∵ (xa)b = xab]
= 2²
= 4

(iii) 16\(\frac{3}{4}\) = (2 × 2 × 2 × 2)\(\frac{3}{4}\)
= (24)\(\frac{3}{4}\)
= 24×\(\frac{3}{4}\) [∵ (xa)b = xab]
= 2³
= 8

Bihar Board Class 9 Maths Solutions Chapter 1 संख्या पद्धति Ex 1.6

(iv) 125\(\frac{-1}{3}\) = (5 × 5 × 5)-\(\frac{1}{3}\)
= (53)-\(\frac{1}{3}\)
= 53×-\(\frac{1}{3}\) [∵ (xa)b = xab]
= 5-1
= \(\frac{1}{5}\) [∵ (x-a) = \(\frac{1}{x^a}\)]

Bihar Board Class 9 Maths Solutions Chapter 1 संख्या पद्धति Ex 1.6

प्रश्न 3.
सरल कीजिए
(i) 2\(\frac{2}{3}\). 2\(\frac{1}{5}\)
(ii) (\(\frac{1}{3³}\))7
(iii) \(\frac{11^\frac{1}{2}}{11^\frac{1}{4}}\)
(iv) 71/2.81/2
हल:
(i) 2\(\frac{2}{3}\). 2\(\frac{1}{5}\) = 2(\(\frac{2}{3}\)+\(\frac{1}{5}\)) [∵ xaxb = xa+b]
= 2\(\frac{10+3}{5}\)
= 213/15

Bihar Board Class 9 Maths Solutions Chapter 1 संख्या पद्धति Ex 1.6

(ii) (\(\frac{1}{3³}\))7 = \(\frac{1}{3^{3×7}}\) [∵ (xa)b = xab]
= \(\frac{1}{3^{21}}\)
= 3-21

(iii) \(\frac{11^\frac{1}{2}}{11^\frac{1}{4}}\) = 11\(\frac{1}{2}\) – \(\frac{1}{4}\) [∵ \(\frac{x^a}{x^b} = x^{a-b}\)]

= 11\(\frac{2-1}{4}\)
= 11\(\frac{1}{4}\)

(iv) 71/2.81/2 = (7.8)\(\frac{1}{2}\) [∵ xaya = (xy)a]
= (56)1/2

Bihar Board Class 9 Maths Solutions Chapter 1 संख्या पद्धति Ex 1.6