Calculati: .....................................

Răspuns
Raspuns - 1\(√2-1)+1\(√2+1)-1\(√3-√2+1\(√3+√2)=
(√2+1)\(2-1)+(√2-1)\(2-1)-(√3+√2)\(3-2)+(√3-√2)\(3-2)=
(√2+1)\1+(√2-1)\1-(√3+√2)\1+(√3-√2)\1=
(√2+√2-1-√3-√2+√3-√2)\1=0\1=0
Explicație pas cu pas:
[tex]\it= \left(\dfrac{1}{\sqrt2-1}+\dfrac{1}{\sqrt2+1}\right)-\left(\dfrac{1}{\sqrt3-\sqrt2}-\dfrac{1}{\sqrt3+\sqrt2}\right)=\dfrac{\sqrt2+1+\sqrt2-1}{(\sqrt2-1)(\sqrt2+1)} -\\ \\ \\ -\dfrac{\sqrt3+\sqrt2-\sqrt3+\sqrt2}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}=\dfrac{2\sqrt2}{1} -\dfrac{2\sqrt2}{1}=0[/tex]