# Rationalize the denominators of the following:

i) 1/√7 ii) 1/(√7 - √6) iii) 1/(√5 + √2) iv) 1/(√7 - 2)

**Solution:**

Rationalization is the process of eliminating a radical or an imaginary number from the denominator of an algebraic fraction.

i) 1/√7

Dividing and multiplying by √7, we get

1/√7 = (1/√7) × (√7/√7)

= √7/7

ii) 1/ (√7 - √6)

Dividing and multiplying by √7 + √6, we get

1/(√7 - √6) = [1/(√7 - √6)] × (√7 + √6) / (√7 + √6)

Using identity (a + b)(a - b) = (a² - b²)

= (√7 + √6) / (√7)² - (√6)²

= (√7 + √6) / (7 - 6)

= √7 + √6

iii) 1/(√5 + √2)

Dividing and multiplying by √5 - √2, we get

= [1/(√5 + √2)] × (√5 - √2)/(√5 - √2)

Using identity (a + b)(a - b) = (a² - b²)

= (√5 - √2) / (√5)² - (√2)²

= (√5 - √2) / (5 - 2)

= (√5 - √2) / 3

iv) 1/(√7 - 2)

Dividing and multiplying by √7 + 2, we get

1/(√7 - 2) = [1/(√7 - 2)] × (√7 + 2)/(√7 + 2)

Using identity (a + b)(a - b) = (a² - b²)

= (√7 + 2) / (√7)² - (2)²

= (√7 + 2) / (7 - 4)

= (√7 + 2) / 3

**☛ Check: **NCERT Solutions for Class 9 Maths Chapter 1

**Video Solution:**

## Rationalize the denominators of the following:

i) 1/√7 ii) 1/ (√7 - √6) iii) 1/ (√5 + √2) iv) 1/(√7 - 2)

NCERT Solutions Class 9 Maths Chapter 1 Exercise 1.5 Question 5:

**Summary:**

Thus, the rationalized expressions of 1/√7, 1/(√7 − √6), 1/(√5 + √2), and 1/(√7 − 2) are √7/7, √7 + √6, (√5 − √2) / 3, and (√7 + 2) / 3 respectively.

**☛ Related Questions:**

- Classify the following numbers as rational or irrational: i) 2 - √5 ii) (3 + √23) - √23 iii) 2√7 ÷ 2√7 iv) 1/√2 v) 2π.
- Simplify each of the following expressions: (i) (3 + √3)(2 + √2) (ii) (3 + √3)(3 - √3) (iii) (√5 + √2)² (iv) (√5 - √2)(√5 + √2)
- Recall, π is defined as the ratio of circumference (say c) of a circle to its diameter (say d). That is, π = c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
- Represent √9.3 on the number line.

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