Question

Write the rationalizing factor of the denominator in 1/root 2+root 2

Answer 1

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↪ Here is your answer :

\begin{lgathered}\sqrt{x} \div \sqrt{2} + \sqrt{2 } \\ \sqrt{x} \div \sqrt{2} + \sqrt{2} \times \sqrt{2} - \sqrt{2} \div \sqrt{2} - \sqrt{2} \\ \sqrt{x} ( \sqrt{2} - \sqrt{2} ) \div ( \sqrt{2} + \sqrt{2} )( \sqrt{2} - \sqrt{2} ) \\ now \: the \: denominator \: becomes \: an \: identity \: which \: is \: (a + b) \: (a - b) = (a ^{2} - b ^{2} ) \\ so \: (\sqrt{2} ^{2} ) - ( \sqrt{2} ^{2} ) = 2 - 2 = 0 \\ so \: \sqrt{x} ( \sqrt{2} - \sqrt{2} ) \div 0\end{lgathered}x÷2+2x÷2+2×2−2÷2−2x(2−2)÷(2+2)(2−2)nowthedenominatorbecomesanidentitywhichis(a+b)(a−b)=(a2−b2)so(22)−(22)=2−2=0sox(2−2)÷0

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