Ad Code

Class 9 Number System Ex 1.5

exercise 1.5 class 9 maths
Question 1:  Find: 

(i) \(64^{\dfrac{1}{2}}\)

Solution:

\(64^{\dfrac{1}{2}}\)

=\({(8 \times 8)}^{\dfrac{1}{2}}\)

=\({{8}^2}^{\times \dfrac{1}{2}}\)

=\(8\)

(ii)\(32^{\dfrac{1}{5}}\)

Solution:

\(32^{\dfrac{1}{5}}\)

=\({(2 \times 2 \times 2 \times 2 \times 2)}^{\dfrac{1}{5}}\)

=\({{2}^5}^{\times \dfrac{1}{5}}\)

=\(2\)

(iii)\(125^{\dfrac{1}{3}}\)

Solution:

\(125^{\dfrac{1}{3}}\)

=\({(5 \times 5 \times 5 )}^{\dfrac{1}{3}}\)

=\({{5}^3}^{\times \dfrac{1}{3}}\)

=\(5\)

Question 2: Find: 

(i)\(9^{\dfrac{3}{2}}\)

Solution:

\(9^{\dfrac{3}{2}}\)

=\({3 \times 2}^{\dfrac{3}{2}}\)

=\({{3}^2}^{\times \dfrac{3}{2}}\)

=\(3^3=27\)

(ii)\(32^{\dfrac{2}{5}}\)

Solution:

\(32^{\dfrac{2}{5}}\)

=\(({2 \times 2 \times 2 \times 2 \times 2})^{\dfrac{2}{5}}\)

=\({{2}^5}^{\times \dfrac{2}{5}}\)

=\(4\)

(iii)\(16^{\dfrac{3}{4}}\)

Solution:

\(16^{\dfrac{3}{4}}\)

=\({(2 \times 2 \times 2 \times 2) }^{\dfrac{3}{4}}\)

=\({{2}^4}^{\times \dfrac{3}{4}}\)

=\(2^3=8\)

(iv)\(125^{\dfrac{-1}{3}}\)
Solution:

\(125^{\dfrac{-1}{3}}\)

=\({(5 \times 5 \times 5) }^{\dfrac{-1}{3}}\)

=\({{5}^3}^{\times \dfrac{-1}{3}}\)

=\(5^{-1}=\dfrac{1}{5}\)

Question 3:

Simplify:

(i) \(2^{\dfrac{2}{3}}.2^{\dfrac{1}{5}}\)

Solution 

\(2^{\dfrac{2}{3}}.2^{\dfrac{1}{5}}\)

= \(2^{{\dfrac{2}{3}}+{\dfrac{1}{5}}}\)

=\(2^{{\dfrac{10+3}{15}}}\)

=\(2^{{\dfrac{13}{15}}}\)

(ii)\(({3^{\dfrac{1}{3}}})^7\)

Solution 

\(\Bigg({3^{\dfrac{1}{3}}}\Bigg)^7\)

=\(\Bigg(3\Bigg)^{\dfrac{7}{3}}\)

(iii)\(\dfrac{11^{\dfrac{1}{2}}}{11^{\dfrac{1}{4}}}\)

Solution 

\(\dfrac{11^{\dfrac{1}{2}}}{11^{\dfrac{1}{4}}}\)

=\(11^{\dfrac{1}{2}-\dfrac{1}{4}}\)

=\(11^{\dfrac{2-1}{4}}\)

=\(11^{\dfrac{1}{4}}\)

(iv) \(7^{\dfrac{1}{2}}.8^{\dfrac{1}{2}}\)

Solution 

\(7^{\dfrac{1}{2}}.8^{\dfrac{1}{2}}\)

=\({(7 \times 8)}^{\dfrac{1}{2}}\)

=\({(56)}^{\dfrac{1}{2}}\)

Post a Comment

0 Comments