Question

Se alguém conseguir resolver, pfvr me ajuda!

Answer 1

a)

${x}^{2} = ({ \sqrt{2} })^{2} + ( { \sqrt{5} })^{2} \\ {x}^{2} = \sqrt{4} + \sqrt{25} \\ {x}^{2} = 2 + 5 \\ {x}^{2} = 7 \\ x = \sqrt{7}$

b)

$({ \sqrt{11} })^{2} = {x}^{2} + ({ \sqrt{5} })^{2} \\ \sqrt{121} = {x}^{2} + \sqrt{25} \\ 11 = {x}^{2} - 5 \\ 11 + 5 = {x}^{2} \\ 16 = {x}^{2} \\ \sqrt{16} = x \\ 4 = x$

c)

$( { \sqrt{7} })^{2} = {x}^{2} + ({ \sqrt{3} })^{2} \\ \sqrt{49} = {x}^{2} + \sqrt{9} \\ 7 = {x}^{2} + 3 \\ 7 - 3 = {x}^{2} \\ 4 = {x}^{2} \\ x = \sqrt{4} \\ x = 2$