Question

11. Rationalize the denominator of
3-√2 by
3+ √2​

Answer 1

Answer:

$\frac{3 - \sqrt{2} }{3 + \sqrt{2} } \times \frac{3 - \sqrt{2} }{3 - \sqrt{2} } \\ \\ = \frac{(3 - \sqrt{2}) {}^{2} }{(3) {}^{2} - ( \sqrt{2) {}^{2} } } \\ \\ = \frac{(3) { }^{2} + ( \sqrt{2}) {}^{2} - 2 \times 3 \times \sqrt{2} }{9 - 2} \\ \\ = \frac{9 - 2 - 6 \sqrt{2} }{9 - 2} \\ \\ = \frac{7 - 6 \sqrt{2} }{7}$

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Answer 2

Answer:

Answer:

\begin{lgathered}\frac{3 - \sqrt{2} }{3 + \sqrt{2} } \times \frac{3 - \sqrt{2} }{3 - \sqrt{2} } \\ \\ = \frac{(3 - \sqrt{2}) {}^{2} }{(3) {}^{2} - ( \sqrt{2) {}^{2} } } \\ \\ = \frac{(3) { }^{2} + ( \sqrt{2}) {}^{2} - 2 \times 3 \times \sqrt{2} }{9 - 2} \\ \\ = \frac{9 - 2 - 6 \sqrt{2} }{9 - 2} \\ \\ = \frac{7 - 6 \sqrt{2} }{7}\end{lgathered}3+23−2×3−23−2=(3)2−(2)2(3−2)2=9−2(3)2+(2)2−2×3×2=9−29−2−62=77−62

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