PSEB 9th Class Maths Solutions Chapter 1 संख्या पद्धति Ex 1.6

Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 1 संख्या पद्धति Ex 1.6 Textbook Exercise Questions and Answers

PSEB Solutions for Class 9 Maths Chapter 1 संख्या पद्धति Ex 1.6

प्रश्न 1.
ज्ञात कीजिए :
(i) \(64^{\frac{1}{2}}\)
(ii) \(32^{\frac{1}{5}}\)
(iii) \(125^{\frac{1}{3}}\)
हल :
(i) \(64^{\frac{1}{2}}\)
= (2 × 2 × 2 × 2 × 2 × 2)^{\(\frac{1}{2}\)}
= \(\left(2^{6}\right)^{\frac{1}{2}}=2^{6 \times \frac{1}{2}}\)
[(a^m)ⁿ = a^{m × n} का प्रयोग करने पर]
= 2³ = 2 × 2 × 2 = 8

(ii) \(32^{\frac{1}{5}}\)
= (2 × 2 × 2 × 2 × 2)^{\(\frac{1}{5}\)}
[(a^m)ⁿ = a^{m × n}] का प्रयोग करने पर
= 2¹ = 2

(iii) \(125^{\frac{1}{3}}\)
= (5 × 5 × 5)^{\(\frac{1}{3}\)}
= \(5^{3 \times \frac{1}{3}}\)
[(a^m)ⁿ = a^{m × n}] का प्रयोग करने पर
= 5¹ = 5

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) \(9^{\frac{3}{2}}\)
= \((3 \times 3)^{\frac{3}{2}}=\left(3^{2}\right)^{\frac{3}{2}}\)
= \(3^{2 \times \frac{3}{2}}\)
[(a^m)ⁿ = a^{m × n} का प्रयोग करने पर]
= 3³ = 3 × 3 × 3 = 27

(ii) \(32^{\frac{2}{5}}\)
= (2 × 2 × 2 × 2 × 2)^{\(\frac{2}{5}\)}
= \(\left(2^{5}\right)^{\frac{2}{5}}=2^{5 \times \frac{2}{5}}\)
[(a^m)ⁿ = a^{m × n} का प्रयोग करने पर]
= 2³ = 2 × 2 = 4

(iii) \(16^{\frac{3}{4}}\)
= (2 × 2 × 2 × 2)^{\(\frac{3}{4}\)}
= \(\left(2^{4}\right)^{\frac{3}{4}}\)
= \(2^{4 \times \frac{3}{4}}\)
[(a^m)ⁿ = a^{m × n} का प्रयोग करने पर]
= 2³ = 2 × 2 × 2 = 8

(iv) \(125^{\frac{-1}{3}}\)
= \((5 \times 5 \times 5)^{\frac{-1}{3}}=\left(5^{3}\right)^{\frac{-1}{3}}\)
= \(5^{3 \times \frac{-1}{3}}\)
[(a^m)ⁿ = a^{m × n} का प्रयोग करने पर]
= 5^-1 = \(\frac{1}{5}\)
[∵ a^-1 = \(\frac{1}{a}\), x ≠ 0]

प्रश्न 3.
सरल कीजिए :
(i) \(2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}\)
(ii) \(\left(\frac{1}{3^{3}}\right)^{7}\)
(iii) \(\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}\)
(iv) \(7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}\)
हल :
(i) \(2^{\frac{2}{3}} \cdot 2^{\frac{1}{5}}\)
= \(2^{\frac{2}{3}+\frac{1}{5}}\)
[a^m . aⁿ = a^m+n का प्रयोग करने पर]
= \(2^{\frac{10+3}{15}}=2^{\frac{13}{15}}\)

(ii) \(\left(\frac{1}{3^{3}}\right)^{7}\)
= (3)^{– 3 × 7}
[(a^m)ⁿ = a^{m × n} का प्रयोग करने पर]
= 3^{– 21}

(iii) \(\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}\)
= \(11^{\frac{1}{2}-\frac{1}{4}}\)
[\(\frac{a^{m}}{a^{n}}\) = a^{m – n} का प्रयोग करने पर]
= \(11^{\frac{2-1}{4}}=11^{\frac{1}{4}}\)

(iv) \(7^{\frac{1}{2}} \cdot 8^{\frac{1}{2}}\)
= (7 × 8)^{\(\frac{1}{2}\)}
[a^m . b^m = (ab)^m का प्रयोग करने पर]
= 56^{\(\frac{1}{2}\)}