Question

a = $\frac{1}{2- \sqrt{3} }$
b = $\frac{1}{2+ \sqrt{3} }$
$(a+b)^{2}$ = ?

Answer 1

$1) A=\frac{1}{2- \sqrt{3} }= \frac{2+ \sqrt{3} }{(2- \sqrt{3} )(2+ \sqrt{3}) }= \frac{2+ \sqrt{3} }{4-3}=2+ \sqrt{3}$
2) B=$\frac{2- \sqrt{3} }{(2+ \sqrt{3})(2- \sqrt{3}) }= \frac{2- \sqrt{3} }{4-3}=2- \sqrt{3}$
La A) si B) s-a folosit formula: $(x+y)(x-y)=x^2-y^2$
3) $(a+b)^2=a^2+2ab+b^2,si,(a-b)^2=a^2-2ab+b^2$