Question

1) Rationalize the denominator.
1/V7+V2​

Answer 1

Given :-

• 1/√7 + √2

To find

• Rationalize its denominator

Solution :-

→ 1/√7 + √2

• Rationalize its denominator by multiplying √7 - √2

→ 1/√7 + √2 × √7 - √2/√7 - √2

• Apply identity
• a² - b² = (a + b)(a - b)

→ √7 - √2/(√7)² - (√2)²

→ √7 - √2/7 - 2

→ √7 - √2/5

Hence,

• The value after rationalising i.e √7 - √2/5

More to know :-

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)

• a³ - b³ = (a - b)(a² + ab + b²)

• a³ + b³ = (a + b)(a² - ab + b²)

Answer 2

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Given :-

• 1/√7 + √2

To do :-

• Rationalize the denominator

How to do:-

• We will rationalize the denominator by using Algebraic Identity.

Algebraic Identity :-

• (a)² - (b)² = (a + b) (a - b)

So, according to the question:-

$\\ \mathtt{\implies { \frac{1}{ \sqrt{7} + \sqrt{2} } }}$

Multiplying √7 + √2 by both numbers

$\\ \mathtt{\implies { {\frac{1}{ \sqrt{7} + \sqrt{2} } \times {\frac{ \sqrt{7} - \sqrt{2} }{ \sqrt{7} - \sqrt{2} } }}}}$

By using identity :-

$\\ \mathtt{ \implies{ \frac{ \sqrt{7} - \sqrt{2} }{ ( \sqrt{7})^{2} - ( \sqrt{2})^{2}} }}$

Therefore,

$\\ \mathtt{ \implies{ \frac{ \sqrt{7} - \sqrt{2} }{7 - 2} }}$

$\\ \color{green} \mathtt{ \implies{ \frac{ \sqrt{7} - \sqrt{2} }{5} }}$

Hence, denominator is in rationalized form.

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