Question

✓3-✓2÷✓3-✓2 rationalise the denominator​

Answer 1

Question :-

$\sf Rationalize\:the\: denominator\:of\: \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}$

Answer :-

In order to rationalize the denominator we have to multiply the fraction with the denominator's inverse such that :

(a-b)(a+b) = a² - b² is formed.

The denominator here = √3 - √2

Inverse = √3 + √2

$\sf Multiplying\: the \:number\: with \: \dfrac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}}$

$\implies \sf \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

$\implies \sf \dfrac{( \sqrt{3} - \sqrt{2})( \sqrt{3} + \sqrt{2} ) }{( \sqrt{3 } - \sqrt{2} )( \sqrt{3} + \sqrt{2}) }$

As we know that (a-b)(a+b) = a² - b², applying the identity,

$\implies \sf \dfrac{( \sqrt{3})^{2} - {( \sqrt{2} )}^{2} }{( \sqrt{3})^{2} - {( \sqrt{2} )}^{2}}$

$\implies \sf \dfrac{3 - 2}{3 - 2}$

$\implies \sf \dfrac{1}{1}$

$\implies \sf 1$

$\sf Therefore\: the\: rarionlized\: form\: of\: \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}\: is\: 1$

Identities :

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• a² - b² = (a - b)(a + b)
• (x + a)(x + b) = x² + x(a + b) + ab
• (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
• (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

Step-by-step explanation:

√3-√2/√3-√2

=1/1

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