Rationalize exercise

Exercise:

Simplify , $\frac{{{\left( \sqrt{2}-1 \right)}^{1-\sqrt{3}}}}{{{\left( \sqrt{2}+1 \right)}^{1+\sqrt{3}}}}$

Solution: we know that $\left( \sqrt{2}+1 \right)\left( \sqrt{2}-1 \right)=2-1=1$

So $\frac{{{\left( \sqrt{2}-1 \right)}^{1-\sqrt{3}}}}{{{\left( \sqrt{2}+1 \right)}^{1+\sqrt{3}}}}\times \frac{{{\left( \sqrt{2}+1 \right)}^{1-\sqrt{3}}}}{{{\left( \sqrt{2}+1 \right)}^{1-\sqrt{3}}}}=\frac{1}{{{\left( \sqrt{2}+1 \right)}^{1+\sqrt{3}+1-\sqrt{3}}}}=\frac{1}{{{\left( \sqrt{2}+1 \right)}^{2}}}$

 $=\frac{1}{{{\left( \sqrt{2}+1 \right)}^{2}}}\times \frac{{{\left( \sqrt{2}-1 \right)}^{2}}}{{{\left( \sqrt{2}-1 \right)}^{2}}}=\frac{{{\left( \sqrt{2}-1 \right)}^{2}}}{{{\left[ \left( \sqrt{2}+1 \right)\left( \sqrt{2}-1 \right) \right]}^{2}}}$

 $={{\left( \sqrt{2}-1 \right)}^{2}}=2+1-2\sqrt{2}=3-2\sqrt{2}$