Question

Sa se aduca la forma mai simpla:
a) 1 supra (1 + radical din 2) + 1 supra (radical din 2 + radical din 3) + 1 supra (radical din 3 + radical din 4).

Answer 1

Salutare!

$\bf \dfrac{1}{1+\sqrt{2}} +\dfrac{1}{\sqrt{2} +\sqrt{3}}+\dfrac{1}{\sqrt{3} +\sqrt{4}} =$

rationalizam radicali cu: (1-√2); (√2-√3); (√3-√4)

$\bf \dfrac{1-\sqrt{2} }{(1+\sqrt{2})\cdot(1-\sqrt{2})} +\dfrac{\sqrt{2} -\sqrt{3}}{(\sqrt{2} +\sqrt{3})\cdot(\sqrt{2} -\sqrt{3})}+\dfrac{\sqrt{3} -\sqrt{4}}{(\sqrt{3} +\sqrt{4})\cdot(\sqrt{3} -\sqrt{4})} =$

$\bf \dfrac{1-\sqrt{2} }{(1^{2}-(\sqrt{2})^{2}} +\dfrac{\sqrt{2} -\sqrt{3}}{(\sqrt{2})^2 -(\sqrt{3})^{2}}+\dfrac{\sqrt{3} -\sqrt{4}}{(\sqrt{3})^{2}-(\sqrt{4})^{2}} =$

$\bf \dfrac{1-\sqrt{2} }{1-2} +\dfrac{\sqrt{2} -\sqrt{3}}{2-3}+\dfrac{\sqrt{3} -\sqrt{4}}{3-4}=$

$\bf -(1-\sqrt{2})-(\sqrt{2} -\sqrt{3})+\dfrac{\sqrt{3} -2}{3-4}=$

$\bf -1+\sqrt{2}-\sqrt{2}+\sqrt{3}-(\sqrt{3} -2)=$

$\bf -1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3} +2=$

$\bf -1+\not\sqrt{2}-\not\sqrt{2}+\not\sqrt{3}-\not\sqrt{3} +2=$

$\bf -1+2=$

$\boxed{\bf 1}$

==pav38==