Question

Dacă a-b=$\sqrt{3}$ și $a^{2}$-$b^{2}$=$\sqrt{6}$, calculați a+b

Answer 1

Răspunsul problemei

Answer 2

$a-b=\sqrt{3}\\a=\sqrt{3} +b\\a^{2} -b^{2} =\sqrt{6}\\(\sqrt{3} +b)^{2} -b^{2} =\sqrt{6} \\3+b^2+2\sqrt{3} b-b^{2} -\sqrt{6} =0\\2\sqrt{3} b=\sqrt{6} -3\\b=\sqrt{6} -3/2\sqrt{3} \\b=(\sqrt{6} -3)\sqrt{3} /6\\b=3\sqrt{2}-3\sqrt{3} /6\\b=\sqrt{2} -\sqrt{3} /2\\a=\sqrt{3} +(\sqrt{2} -\sqrt{3} )/2\\a=(\sqrt{3}+\sqrt{2} )/2\\p:(\sqrt{3} +\sqrt{2}-\sqrt{2} +\sqrt{3})/2 =2 \sqrt{3} /2 = \sqrt{3}$

$a+b= \sqrt{2}$