CALCULER (2-[3])^ - ([5]-1)^ = ?? [...] = racine carré ^ = exposant 2

Salut,

[tex](2- \sqrt{3})^{2} - ( \sqrt{5} -1)^{2} \\ 2^{2} - 2*2* \sqrt{3} + \sqrt{3}^{2} - ( \sqrt{5} ^{2} - 2 * 1 * \sqrt{5} +1^{2}) \\ 4 - 4\sqrt{3} + 3 - (5 -2 \sqrt{5} +1) \\ 7 - 4\sqrt{3} - 5 + 2 \sqrt{5} - 1 \\ 1 - 4\sqrt{3} + 2 \sqrt{5}[/tex]

Bonne soirée !

Bonsoir,

[tex](2-\sqrt{3})^2 - (\sqrt{5}-1)^2\\\\=(2^2-2\times2\times\sqrt{3}+(\sqrt{3})^2) -((\sqrt{5})^2-2\times\sqrt{5}\ti+1^2)\\\\=(4-4\sqrt{3}+3) -(5-2\sqrt{5}+1)\\\\=(7-4\sqrt{3}) -(6-2\sqrt{5})\\\\=7-4\sqrt{3} -6+2\sqrt{5}\\\\=1-4\sqrt{3} +2\sqrt{5}[/tex]