Question

Rationalize the denominator of 3-√2/3+√2

Answer 1

Answer:

= $\frac{13-6\sqrt{2}}{5}$

Step-by-step explanation:

here, $\frac{3-\sqrt{2} }{3+\sqrt{2} }$

      = $\frac{3-\sqrt{2} }{3+\sqrt{2} }$ × $\frac{3-\sqrt{2} }{3-\sqrt{2} }$      by rationalization

      = $\frac{(3-\sqrt{2})(3-\sqrt{2})}{(3+\sqrt{2})(3-\sqrt{2}) }$

      = $\frac{9-3\sqrt{2}-3\sqrt{2}+4}{9-4}$

      = $\frac{13-6\sqrt{2}}{5}$

so, rationalized denominator is = $\frac{13-6\sqrt{2}}{5}$

hope it will help you.