Question

como eu posso simplificar esses dois radicais e reduzir os termo ??
 A) 5√2 - 2√5 + √50 - 2√20 + √500:

 B) -1 √44 + 3√99 - 2 √11 - 1√1100:
      2                     3         4                                                                                                               
agradeço a quem poder me ajudar!!  :)     

Answer 1

$10 \sqrt{2} +6 \sqrt{5}$

Answer 2

$\sqrt{50} = 5*5*2 = 5 \sqrt{2}\\\\\ \sqrt{20} = 2*2*5 = 2 \sqrt{5} *2 = 4 \sqrt{5} \\\\\ \sqrt{500} = 5*5*2*2*5 =10 \sqrt{5}\\\\\\ 5 \sqrt{2} -2 \sqrt{5} +5 \sqrt{2} -4 \sqrt{5} +10 \sqrt{5} \\\\\\ \boxed{10 \sqrt{2} +4 \sqrt{5} }$

B)

$\sqrt{44} = 2*2*11 = 2 \sqrt{11} *\frac{1}{2} = \sqrt{11} \\\\\ \sqrt{99} = 3*3*11 = 3 \sqrt{11} *3 = 9\sqrt{11}\\\\\ \sqrt{1100} = 10\sqrt{11}\\\\\\\ -\sqrt{11}+9\sqrt{11}-\frac{2}{3}\sqrt{11}-\frac{5}{2}\sqrt{11}\\\\\\\ -6\sqrt{11}+54\sqrt{11}-4\sqrt{11}-15\sqrt{11}$

${2}\sqrt{11}\\\\\\\ -6\sqrt{11}+54\sqrt{11}-4\sqrt{11}-15\sqrt{11} = \frac{29 \sqrt{11} }{6}\\\\\ \boxed{ \frac{29 \sqrt{11} }{6}}$