Math, asked by majidkhan2493, 7 months ago


rationalise \: the \: denomiaor \: of \:  \sqrt{} 3 +  \sqrt{} 2 \div  \sqrt{} 3 -  \sqrt{} 2

Answers

Answered by aryan073
7

Solution ::

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\huge\boxed{\fcolorbox{red}{Aqua}{Answer}}

By rationalizing them from numerator and denominator u get this.

\\    \red\bigstar\tt\:  \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}

\\    \red\bigstar\tt\:  \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}  \times  \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}      \times   \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}

\\    \red\bigstar\tt\:  \dfrac{3-2}{3-2} \times \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}[/tex]

\\    \red\bigstar\tt\:  \dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}

Now rationalizing them by the denominator :

\\    \red\bigstar\tt\:  \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}

\\    \red\bigstar\tt\:  \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} \times \dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}[/tex]

\\    \red\bigstar\tt\:  \dfrac{(\sqrt{3}+\sqrt{2})^2}{3-2}

\\    \red\bigstar\tt\:  \dfrac{(\sqrt{3}+\sqrt{2})^2}{1}

\\    \red\bigstar\tt\:  \dfrac{\sqrt{3}+\sqrt{2}}{1}

\\    \red\bigstar\tt\:  \dfrac{3+2+2\sqrt{6}}{1}

\\    \red\bigstar\tt\:  \dfrac{5+2\sqrt6}{1}

\huge\mathfrak\pink{Solution :-}

\sf{5+2\sqrt6}

\bold{\purple{\fbox{\red{Please thank it}}}}

Answered by rk4846336
12

Answer:

  \frac{ \sqrt[]{3}  +  \sqrt[]{2} }{ \sqrt[]{3}  -  \sqrt[]{2} }   \\  = \frac{ \sqrt[]{3}  +  \sqrt[]{2} }{ \sqrt[]{3}  -  \sqrt[]{2} } \times \frac{ \sqrt[]{3}    +  \sqrt[]{2} } { \sqrt[]{3}    + \sqrt[]{2} } \\  = \frac{ (\sqrt[]{3}  +  \sqrt[]{2} {)}^{2}  }{ (\sqrt[]{3} ) {}^{2}  -  (\sqrt[]{2} ) {}^{2} } {}^{}  \\  \\  =  \frac{( \sqrt[]{3} ) {}^{2}  +  \sqrt[]({2} ) {}^{2} + 2 \times  \sqrt[]{3}   \times \sqrt[]{2}  }{3 - 2} \\   = \frac{3 + 2 + 2 \sqrt[]{6} }{1 }  \\  = 5 + 2 \sqrt{6}

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