Correction de

1. Calculons A+B.
$A=2-\sqrt{5}$ et $B=1+2\sqrt{5}$

$A+B=2-\sqrt{5}+1+2\sqrt{5}$

$\iff$ $A+B=1+2-\sqrt{5}+2 \sqrt{5}$

$\iff$ $A+B=3+\sqrt{5}$
1. Justifions que $A\times B=C$
$A\times B=(2-\sqrt{5})\times(1+2\sqrt{5})$

$A\times B=2\times1+2\times2\sqrt{5}-\sqrt{5}\times1+\sqrt{5}\times2\sqrt{5}$

$A\times B=2+4\sqrt{5}-2\sqrt{5}\times\sqrt{5}$

$A\times B=2-4\sqrt{5}+10$

$A\times B=-8+4\sqrt{5}$
1. Ecrivons $\frac{1}{A}$
$\frac{1}{A}=\frac{1}{2-\sqrt{5}}$<br\>$\iff$ $\frac{1}{A}=\frac{2+\sqrt{5}}{(2-\sqrt{5})(2+\sqrt{5})}$<br\> $\iff$ $\frac{1}{A}=\frac{2+\sqrt{5}}{2^2-(\sqrt{5})^2}$<br\>$\iff$ $\frac{1}{A}=\frac{2+\sqrt{5}}{4-5}$

$\iff$ $\underline{\frac{1}{A}=-2-\sqrt{5}}$