Question

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

Answer 1

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:

Answer 2

$\it= \left(\dfrac{1}{\sqrt2-1}+\dfrac{1}{\sqrt2+1}\right)-\left(\dfrac{1}{\sqrt3-\sqrt2}-\dfrac{1}{\sqrt3+\sqrt2}\right)=\dfrac{\sqrt2+1+\sqrt2-1}{(\sqrt2-1)(\sqrt2+1)} -\\ \\ \\ -\dfrac{\sqrt3+\sqrt2-\sqrt3+\sqrt2}{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}=\dfrac{2\sqrt2}{1} -\dfrac{2\sqrt2}{1}=0$