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Rationalise the denominator is 1/(√4 - √2)​

Sagot :

Answer:

[tex] = { \tt{ \frac{1}{( \sqrt{4} - \sqrt{2} )}. \frac{ \sqrt{4} + \sqrt{2} }{ \sqrt{4} + \sqrt{2} } }} \\ \\ = { \tt{ \frac{ \sqrt{4} + \sqrt{2} }{( \sqrt{4}) {}^{2} - {( \sqrt{2}) }^{2} } }} \\ \\ = { \tt{ \frac{ \sqrt{4} + \sqrt{2} }{4 - 2} }} \\ \\ = { \tt{ \frac{2 + \sqrt{2} }{2} } } \\ \\ = { \boxed{ \tt{ \: \: 1 + \frac{ \sqrt{2} }{2} }}}[/tex]

SalmonWorldTopper

Step-by-step explanation:

[tex]\underline{\underline{\sf{➤\:\: Solution }}}[/tex]

[tex] \sf(a) \: \: \: \dfrac{1}{ \sqrt{4} - \sqrt{2} } [/tex]

On rationalising,

[tex] \sf \implies \dfrac{1}{ \sqrt{4} - \sqrt{2} } \times \dfrac{\sqrt{4} + \sqrt{2} }{\sqrt{4} + \sqrt{2} } [/tex]

Combine the fractions,

[tex] \sf \implies \dfrac{1(\sqrt{4} + \sqrt{2}) }{(\sqrt{4} - \sqrt{2})(\sqrt{4} + \sqrt{2}) } [/tex]

We know that,

[tex] \sf \implies (a - b)(a + b) = (a)^{2} - (b)^{2} [/tex]

So,

[tex] \sf \implies \dfrac{1(\sqrt{4} + \sqrt{2}) }{(\sqrt{4})^{2} - (\sqrt{2}) ^{2} }[/tex]

[tex] \sf \implies \dfrac{1(\sqrt{4} + \sqrt{2}) }{4 - 2 }[/tex]

[tex] \sf \implies \dfrac{1(\sqrt{4} + \sqrt{2}) }{2 }[/tex]

[tex] \sf \implies \dfrac{\sqrt{4} + \sqrt{2}}{2} [/tex]

Hence,

Hence, On rationalising we got,

[tex]\implies \bf {\dfrac{\sqrt{4} + \sqrt{2}}{2}} [/tex]

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