Bonsoir,
6)
[tex]A = (3 - \sqrt{5})(3 + \sqrt{5})\\\\A = 3^2 - \sqrt{5}^2\\\\A = 9 - 5\\\\\boxed{A = 4}[/tex]
C'est une identité remarquable (a - b)(a + b) = a² - b².
Elle se retrouve en faisant une double distributivité.
7)
[tex]B = \sqrt{18}\\\\B = \sqrt{9\cdot 2}\\\\B = \sqrt{9}\sqrt{2}\\\\B = \sqrt{3^2}\sqrt{2}\\\\\boxed{B = 3\sqrt{2}}[/tex]
8)
[tex]C = -2\sqrt{12}\\\\C = -2\sqrt{4\cdot 3}\\\\C = -2\sqrt{4}\sqrt{3}\\\\C = -2\sqrt{2^2}\sqrt{3}\\\\C = -2\cdot2\cdot\sqrt{3}\\\\\boxed{C = -4\sqrt{3}}[/tex]
9)
[tex]D = 6\sqrt{3} - \sqrt{12} - \sqrt{27}\\\\D = 6\sqrt{3} - \sqrt{4\cdot 3} - \sqrt{9\cdot 3}\\\\D = 6\sqrt{3} - \sqrt{4}\sqrt{3} - \sqrt{9}\sqrt{3}\\\\D = 6\sqrt{3} - \sqrt{2^2}\sqrt{3} - \sqrt{3^2}\sqrt{3}\\\\D = 6\sqrt{3} - 2\sqrt{3} - 3\sqrt{3}\\\\D = (6-2-3)\sqrt{3}\\\\\boxed{D = \sqrt{3}}[/tex]
10)
[tex]E = \sqrt{5} \cdot 4\sqrt{5} - \sqrt{\frac{1}{25}}\cdot 15\\\\E = 4\cdot\sqrt{5}^2 - \frac{\sqrt{1}}{\sqrt{25}}\cdot 15\\\\E = 4 \cdot 5 - \frac{1}{\sqrt{5^2}} \cdot 5\cdot 3\\\\E = 20 - \frac{5\cdot 3}{5}\\\\E = 20 - 3\\\\\boxed{E = 17}[/tex]
Bonne soirée,
Thomas
