Bonjour,
[tex]A=\dfrac{\sqrt{5}}{1+\sqrt{5}}\\\\A=\dfrac{\sqrt{5}(1-\sqrt{5})}{(1+\sqrt{5})(1-\sqrt{5})}\\\\A=\dfrac{\sqrt{5}-(\sqrt{5})^2}{1^2-(\sqrt{5})^2}\\\\A=\dfrac{\sqrt{5}-5}{1-5}\\\\A=\dfrac{\sqrt{5}-5}{-4}\\\\\boxed{A=\dfrac{5-\sqrt{5}}{4}}[/tex]
[tex]B=\dfrac{\sqrt{2}-\sqrt{7}}{5-\sqrt{8}}\\\\B=\dfrac{(\sqrt{2}-\sqrt{7})(5+\sqrt{8})}{(5-\sqrt{8})(5+\sqrt{8})}\\\\B=\dfrac{5\sqrt{2}+\sqrt{2}\sqrt{8}-5\sqrt{7}-\sqrt{7}\sqrt{8}}{5^2-(\sqrt{8})^2}\\\\B=\dfrac{5\sqrt{2}+\sqrt{16}-5\sqrt{7}-\sqrt{56}}{25-8}\\\\B=\dfrac{5\sqrt{2}+4-5\sqrt{7}-\sqrt{4\times14}}{17}\\\\B=\dfrac{5\sqrt{2}+4-5\sqrt{7}-2\sqrt{14}}{17}\\\\\boxed{B=\dfrac{4+5\sqrt{2}-5\sqrt{7}-2\sqrt{14}}{17}}[/tex]
