Question

$\sqrt{(6- \sqrt{5 } } ^{2} + \sqrt{41+12 \sqrt{5} } ^{2} ori x = \sqrt{(6- \sqrt{5} } ^{2} ori x - \sqrt{41-12 \sqrt{5} }$ rezolvare completa.

Answer 1

$\sqrt{(6-\sqrt{5})}^2+\sqrt{41+2\sqrt5}^2\cdot x=\sqrt{(6-\sqrt{5})}^2\cdot x -\sqrt{41-2\sqrt5} \\ \\ |6-\sqrt5|+|41+2\sqrt5|\cdot x = |6-\sqrt5|\cdot x - \sqrt{41-2\sqrt5} \\ \\ 6-\sqrt5+ (41+2\sqrt5)\cdot x = (6-\sqrt5)\cdot x-\sqrt{41-2\sqrt5} \\ \\ (41+2\sqrt5)\cdot x- (6-\sqrt5)\cdot x = -(6-\sqrt5)- \sqrt{41-2\sqrt5} \\ \\ x\cdot \Big(41+2\sqrt5-6+\sqrt5\Big) = -6+\sqrt5-\sqrt{41-2\sqrt5}$

$x\cdot (35+3\sqrt5) = \sqrt5-\sqrt{41-2\sqrt5}-6 \\ \\ x =^{\Big{35-3\sqrt5)}}\dfrac{\sqrt5-\sqrt{41-2\sqrt5}-6}{35+3\sqrt5} \\ \\ \\ x = \dfrac{(35-3\sqrt5)\cdot ({\sqrt5-\sqrt{41-2\sqrt5}-6)}}{35^2-45}\\ \\ x = \dfrac{(35-3\sqrt5)\cdot ({\sqrt5-\sqrt{41-2\sqrt5}-6)}}{1180}$