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Ex clasa a 8-a, mulțumesc ​

Ex Clasa A 8a Mulțumesc class=

Răspuns :

[tex](\frac{1}{\sqrt5}+1)^2+(\frac{1}{\sqrt5}-1)^2-\frac{3}{5}=(\frac{\sqrt5}{5}+1)^2+(\frac{\sqrt5}{5}-1)^2-\frac{3}{5}=(\frac{\sqrt5}{5})^2+\frac{2\sqrt5*1}{5}+1^2}+(\frac{\sqrt5}{5})^2-\frac{2\sqrt5*1}{5}+1^2}-\frac{3}{5}=\frac{5}{25}+1+\frac{5}{25}+1-\frac{3}{5}=\frac{10}{25}+\frac{10}{5}-\frac{3}{5}=\frac{2}{5}+\frac{7}{5}=\frac{9}{5}\\[/tex]