Question

please answer this question​

Answer 1

1/1+3√2

=1-3√2/(1+3√2)(1-3√2)

=1-3√2/1²-(3√2)²

=1-3√2/1-3²(√2)²

=1-3√2/1-9(√2)²

=1-3√2/1-18

=1-3√2/-17

=-1+3√2/17

hope it is helpful to you

Answer 2

Step-by-step explanation:

Rationalizing factor of $\bold{\frac{1}{1+3\sqrt{2} } }$ ?

$\rightarrowtail \bold{\frac{1}{1+3\sqrt{2}\ } \bold{\times}\frac{1-3\sqrt{2} }{1-3\sqrt{2} }}$

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

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

$\rightarrow \bold{1-(3\sqrt{2})^2= 1 - 3\sqrt{2}\times3\sqrt{2} }$

$\rightarrow \bold{1-3\times3\times\sqrt{2}\times\sqrt{2} }$

$\rightarrow \bold{1-9\times\sqrt{4} }$

$\rightarrow \bold{1 - 9\times 2}$

$\rightarrow \bold{1-18=-17}$

$\rightarrowtail \bold{\frac{1-3\sqrt{2} }{-17}\;or\;\frac{1+3\sqrt{2} }{17} }$

∴ $\bold{\frac{1+3\sqrt{2} }{17} }$ is the rationalizing factor of $\bold{\frac{1}{1+3\sqrt{2} } }$.