Rozwiązane

jakby ktos mogl rozpisac bylbym wdzieczny :D[tex]\sqrt[]{2(2-2\sqrt2)^{2}} + \sqrt{2(2+2\sqrt2)^{2}[/tex]



Odpowiedź :

Konrad509

[tex]\begin{aligned}&\sqrt{2(2-2\sqrt2)^2}+\sqrt{2(2+2\sqrt2)^2}=\\&\sqrt2\cdot\sqrt{(2-2\sqrt2)^2}+\sqrt2\cdot \sqrt{(2-2\sqrt2)^2}=\\&\sqrt 2\cdot |2-2\sqrt2|+\sqrt 2\cdot |2+2\sqrt2|=\\&\sqrt 2\cdot (-2+2\sqrt2)+\sqrt2\cdot (2+2\sqrt2)=\\&-2\sqrt2+2\cdot2+2\sqrt2+2\cdot2=\\&4+4=8\end[/tex]

Basetla

[tex]\sqrt{2(2-2\sqrt{2})^{2}}+\sqrt{2(2+2\sqrt{2})^{2}}=\sqrt{2}\cdot\sqrt{(2-2\sqrt{2})^{2}}+\sqrt{2}\cdot\sqrt{(2+2\sqrt{2})^{2}}=\\\\=\sqrt{2}\cdot|2-2\sqrt{2}|+\sqrt{2}\cdot|2+2\sqrt{2}|=\sqrt{2}(2\sqrt{2}-2)+\sqrt{2}(2+2\sqrt{2})=\\\\=\sqrt{2}(2\sqrt{2}-2+2+2\sqrt{2}) = \sqrt{2}\cdot4\sqrt{2} = 8[/tex]