Math, asked by Anonymous, 1 year ago

rationalize the denominator: 1/1+root2-root3

Answers

Answered by Aloi99
4

Question:-

✪Rationalize  \frac{1}{1+\sqrt{2}-\sqrt{3}}

\rule{200}{1}

Answer:-

 \frac{\sqrt{2}-\sqrt{3}}{6}

\rule{200}{1}

Proof:-

 \frac{1}{1+\sqrt{2}-\sqrt{3}}

•Multiply both Numerator and Denominator by 1-√2-√3•

 \frac{1\times (1-\sqrt{2}-\sqrt{3})}{1+\sqrt{2}-\sqrt{3}\times (1-\sqrt{2}-\sqrt{3})}

★Applying a²-b²=(a+b)(a-b) identity★

 \frac{1-\sqrt{2}-\sqrt{3}}{(1)^{2}-(\sqrt{2}-\sqrt{3})^{2}}

 \frac{1-\sqrt{2}-\sqrt{3}}{1-(4-9)}

 \frac{1-\sqrt{2}-\sqrt{3}}{1-(-5)}

 \frac{1-\sqrt{2}-\sqrt{3}}{1+5}

 \frac{1-\sqrt{2}-\sqrt{3}}{6}

\rule{200}{2}

Answered by Saby123
5

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QUESTION -

rationalize the denominator: 1/1+root2-root3.

SOLUTION -

 \dfrac{1}{ 1 + \sqrt{2} - \sqrt{3} } \\ \\ => \dfrac{1}{ 1 + \sqrt{2} - \sqrt{3} } \times \dfrac{1 + \sqrt{2} + \sqrt{3} }{ 1 + \sqrt{2} + \sqrt{3} } \\ \\=>  \dfrac{1 + \sqrt{2} + \sqrt{3} }{  \sqrt{2} - 6 } \\ \\

 => \dfrac{1 + \sqrt{2} + \sqrt{3} }{  \sqrt{2} - 6 } \times \dfrac{ \sqrt{2} + 6 }{ \sqrt{2 } + 6} \\ \\ => \dfrac{1 - \sqrt{2} - \sqrt{3}  }{6  }

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