Rationalizing and Solving problems

Rationalize the following :-

1). 1/√7 = (1/√7) * (√7 / √7) = (√7 / 7)

2). 1/√5 = (1/√5) * (√5 / √5) = (√5 / 5)

3). 1/ (√5 + √2) = {1/ (√5 + √2)} * {( √5 - √2 ) / (√5 - √2)] = { (√5 - √2) / (5 – 2)} = (√5 - √2) / 3

4). 1/ (√5 - √2) = {1/ (√5 - √2)}*{( √5 + √2 ) / (√5 + √2)] = { (√5 + √2) / (5 – 2)} = (√5 + √2) / 3

5). 1/ (√5 + √3) = {1/ (√5 + √3)} * {( √5 - √3 ) / (√5 - √3)] = { (√5 - √3) / (5 – 3)} = (√5 - √3) / 2

6). 1/ (√3 + √2) = {1/ (√3 + √2)} * {( √3 - √2 ) / (√3 - √2)] = { (√3 - √2) / (3 – 2)} = (√3 - √2)

7). 1/ (√5 - √3) = {1/ (√5 - √3)} * {( √5 + √3 ) / (√5 + √3)] = { (√5 + √3) / (5 – 3)} = (√5 + √3) / 2

8). 1/ (√3 - √2) = {1/ (√3 - √2)} * {( √3 + √2 ) / (√3 + √2)] = { (√3 - √2) / (3 – 2)} = (√3 - √2)

9). 1/ (√7 + 2) = {1/ (√7 + 2)} * {( √7 - 2 ) / (√7 - 2)] = { (√7 - 2) / (7 – 4)} = (√7 - 2) / 3

10). 1/ (√7 - 2) = {1/ (√7 - 2)} * {( √7 + 2 ) / (√7 + 2)] = { (√7 + 2) / (7 – 4)} = (√7 + 2) / 3

11). 1/ (√7 - √6) = {1/ (√7 - √6)} * {( √7 + √6 ) / (√7 + √6)] = { (√7 + √6) / (7 – 6)} = (√7 + √6)

Note :-

Use conjugate to rationalize numbers.

maths-ncert-solution

Solve the following ;-

1). (64) ^ (1/2) = (2^6)^(1/2) = 2^3 = 8 { (x^m)^n = x^(m*n)}

2). (16) ^ (1/2) = (2^4)^(1/2) = 2^2 = 4 { (x^m)^n = x^(m*n)}

3). (32) ^ (1/5) = (2^5)^(1/5) = 2^1 = 2 { (x^m)^n = x^(m*n)}

4). (128) ^ (1/7) = (2^7)^(1/7) = 2^1 = 2 { (x^m)^n = x^(m*n)}

5). (25) ^ (1/2) = (5^2)^(1/2) = 5^1 = 5 { (x^m)^n = x^(m*n)}

6). (125) ^ (1/3) = (5^3)^(1/3) = 5^1 = 5 { (x^m)^n = x^(m*n)}

7). (625) ^ (1/2) = (5^4)^(1/2) = 5^2 = 25 { (x^m)^n = x^(m*n)}

8). (27) ^ (1/3) = (3^3)^(1/3) = 3^1 = 3 { (x^m)^n = x^(m*n)}

9). (81) ^ (1/2) = (3^4)^(1/2) = 3^2 = 9 { (x^m)^n = x^(m*n)}

10). (243) ^ (1/5) = (3^5)^(1/5) = 3^1 = 3 { (x^m)^n = x^(m*n)}

11). (49) ^ (1/2) = (7^2)^(1/2) = 7^1 = 7 { (x^m)^n = x^(m*n)}

12). (64) ^ (3/2) = (2^6)^(3/2) = 2^9 = 1024 { (x^m)^n = x^(m*n)}

13). (16) ^ (-1/2) = (2^4)^(-1/2) = 2^(-2) = 1/4 { (x^m)^n = x^(m*n)}

14). (32) ^ (3/5) = (2^5)^(3/5) = 2^3 = 8 { (x^m)^n = x^(m*n)}

15). (128) ^ (2/7) = (2^7)^(2/7) = 2^2 = 4 { (x^m)^n = x^(m*n)}

16). (25) ^ (3/2) = (5^2)^(3/2) = 5^3 = 125 { (x^m)^n = x^(m*n)}

17). (125) ^ (2/3) = (5^3)^(2/3) = 5^2 = 25 { (x^m)^n = x^(m*n)}

18). (625) ^ (3/4) = (5^4)^(3/4) = 5^3 = 125 { (x^m)^n = x^(m*n)}

19). (27) ^ (-2/3) = (3^3)^(-2/3) = 3^(-2) = 1/9 { (x^m)^n = x^(m*n)}

20). (81) ^ (-3/2) = (3^4)^(-3/2) = 3^(-6) = 1/729 { (x^m)^n = x^(m*n)}

21). (243) ^ (-1/5) = (3^5)^(-1/5) = 3^(-1) = 1/3 { (x^m)^n = x^(m*n)}

22). (49) ^ (-1/2) = (7^2)^(-1/2) = 7^(-1) = 1/7 { (x^m)^n = x^(m*n)}

23). [{2^(2/3)} * {2^(1/5)}] = 2^(2/3 + 1/5) = 2^{(10+3)/15} = 2^(13/15) [(x^a) * (x^b) = x^(a + b)]

24). [{3^(2/3)} * {3^(1/5)}] = 3^(2/3 + 1/5) = 3^{(10+3)/15} = 3^(13/15) [(x^a) * (x^b) = x^(a + b)]

25). [{5^(2/3)} * {5^(1/5)}] = 5^(2/3 + 1/5) = 5^{(10+3)/15} = 5^(13/15) [(x^a) * (x^b) = x^(a + b)]

26). {1/(3^3)}^7 = {3^(-3)}^7 = 3^(-21) (x^m)^n = x^(m*n)

27). {1/(2^3)}^7 = {2^(-3)}^7 = 2^(-21) (x^m)^n = x^(m*n)

28). {1/(5^3)}^7 = {5^(-3)}^7 = 5^(-21) (x^m)^n = x^(m*n)

29). (2^1/2) / 2^1/4) = 2^(1/2 – 1/4) = 2^(1/4) [(x^a) / (x^b) = x^(a - b)]

30). (3^1/2) / 3^1/4) = 3^(1/2 – 1/4) = 3^(1/4) [(x^a) / (x^b) = x^(a - b)]

31). (5^1/2) / 5^1/4) = 5^(1/2 – 1/4) = 5^(1/4) [(x^a) / (x^b) = x^(a - b)]

32). (2^1/2) * (3^1/2) = (2*3)^(1/2) = 6^(1/2) [(x^a) * (y^a) = (x*y)^a]

33). (2^1/3) * (3^1/3) = (2*3)^(1/3) = 6^(1/3) [(x^a) * (y^a) = (x*y)^a]

34). (2^1/4) * (3^1/4) = (2*3)^(1/4) = 6^(1/4) [(x^a) * (y^a) = (x*y)^a]

35). (2^1/5) * (3^1/5) = (2*3)^(1/5) = 6^(1/5) [(x^a) * (y^a) = (x*y)^a]

Note :-

* [(x^m)^n = x^(m*n)]

* [(x^a) * (x^b) = x^(a + b)]

* [(x^a) / (x^b) = x^(a - b)]

* [(x^a) * (y^a) = (x*y)^a]

Milan Tomic

Hi. I’m Designer of Blog Magic. I’m CEO/Founder of ThemeXpose. I’m Creative Art Director, Web Designer, UI/UX Designer, Interaction Designer, Industrial Designer, Web Developer, Business Enthusiast, StartUp Enthusiast, Speaker, Writer and Photographer. Inspired to make things looks better.