Bonjour à tous Je suis bloqué sur cet exercice. Merci de détailler les calculs svp.

Bonjour MrMaël

[tex]a)\ -2x^2+(2\sqrt{5}-1)x+\sqrt{5}=0\\\Delta=(2\sqrt{5}-1)^2-4\times(-2)\times\sqrt{5}\\\Delta=(2\sqrt{5})^2-2\times2\sqrt{5}\times1+1^2+8\sqrt{5}\\\Delta=4\times5-4\sqrt{5}+1+8\sqrt{5}\\\Delta=20+1+4\sqrt{5}\\\Delta=21+4\sqrt{5}\ \textgreater \ 0\\\\x_1=\dfrac{-(2\sqrt{5}-1)-\sqrt{21+4\sqrt{5}}}{2\times(-2)}\\\\x_1=\dfrac{-2\sqrt{5}+1-\sqrt{21+4\sqrt{5}}}{-4}[/tex]

[tex]\\\\\boxed{x_1=\dfrac{2\sqrt{5}-1+\sqrt{21+4\sqrt{5}}}{4}}[/tex]

[tex]\\\\x_2=\dfrac{-(2\sqrt{5}-1)+\sqrt{21+4\sqrt{5}}}{2\times(-2)}\\\\x_2=\dfrac{-2\sqrt{5}+1+\sqrt{21+4\sqrt{5}}}{-4}}\\\\\boxed{x_2=\dfrac{2\sqrt{5}-1-\sqrt{21+4\sqrt{5}}}{4}}}[/tex]

[tex]b)\ (1+2\sqrt{5})^2=1^2+2\times1\times2\sqrt{5}+(2\sqrt{5})^2\\\\(1+2\sqrt{5})^2=1+4\sqrt{5}+4\times5\\\\(1+2\sqrt{5})^2=1+4\sqrt{5}+20\\\\\boxed{(1+2\sqrt{5})^2=21+4\sqrt{5}}[/tex]

[tex]c)\ -2x^2+(2\sqrt{5}-1)x+\sqrt{5}=0\\\\\Delta=21+4\sqrt{5}\ \textgreater \ 0\\\sqrt{\Delta}=\sqrt{21+4\sqrt{5}}=\sqrt{(1+2\sqrt{5})^2}=1+2\sqrt{5}\\\\x_1=\dfrac{2\sqrt{5}-1+\sqrt{21+4\sqrt{5}}}{4}\\\\x_1=\dfrac{2\sqrt{5}-1+1+2\sqrt{5}}{4}\\\\x_1=\dfrac{4\sqrt{5}}{4}\\\\\boxed{x_1=\sqrt{5}}[/tex]

[tex]x_2=\dfrac{2\sqrt{5}-1-\sqrt{21+4\sqrt{5}}}{4}\\\\x_2=\dfrac{2\sqrt{5}-1-(1+2\sqrt{5})}{4}}\\\\x_2=\dfrac{2\sqrt{5}-1-1-2\sqrt{5}}{4}}\\\\x_2=\dfrac{-2}{4}}\\\\\boxed{x_2=\dfrac{-1}{2}}[/tex]