Question

Buna ziua! Imi poate spune cineva, va rog, urgent, cu cat este egal b? Si as vrea si rezolvarea, daca se poate. Multumesc!!!

Answer 1

$\bf b = \dfrac{\sqrt{2}+\sqrt{3}+\sqrt{32}+\sqrt{48}}{\sqrt{2}-\sqrt{3}+\sqrt{32}-\sqrt{48}} :\dfrac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}$

$\bf b = \dfrac{\sqrt{2}+\sqrt{3}+4\sqrt{2}+4\sqrt{3}}{\sqrt{2}-\sqrt{3}+4\sqrt{2}-4\sqrt{3}} :\dfrac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}$

$\bf b = \dfrac{5\sqrt{2}+5\sqrt{3}}{5\sqrt{2}-5\sqrt{3}} \cdot\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}$

$\bf b = \dfrac{5\cdot(\sqrt{2}+\sqrt{3})}{5\cdot(\sqrt{2}-\sqrt{3})} \cdot\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}$

$\bf Facem ~simplificarile ~corespunzatoare$

$\bf b = \dfrac{5\cdot\not(\sqrt{2}+\sqrt{3})}{5\cdot\not(\sqrt{2}-\sqrt{3})} \cdot\dfrac{\not(\sqrt{2}-\sqrt{3})}{\not(\sqrt{2}+\sqrt{3})}$

$\bf b = \dfrac{5}{5} \cdot\dfrac{1}{1}$

$\bf b = \dfrac{\not5}{\not 5} \cdot\dfrac{1}{1}$

$\bf b = \dfrac{1}{1} \cdot\dfrac{1}{1}$

$\red{\large \boxed{\bf b = 1}}$

Bafta multa !