Question

mim ajudar a responde por favor estou precisando muito muito mesmo​

Answer 1

Resposta:

Explicação passo a passo:

a)  $\frac{1}{3-\sqrt{6} } . \frac{3+\sqrt{6}}{3+\sqrt{6}} = \frac{3+\sqrt{6}}{9-6} =\frac{3+\sqrt{6}}{3}$

b)  $\frac{2}{\sqrt{5}+\sqrt{3}} . \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}} = \frac{2\sqrt{5}-2\sqrt{3}}{5-3} = \frac{2\sqrt{5}-2\sqrt{3}}{2}=\frac{2(\sqrt{5}-\sqrt{3})}{2}=\sqrt{5}-\sqrt{3}$

c) $\frac{11}{2\sqrt{3}-1} . \frac{2\sqrt{3}+1}{2\sqrt{3}+1}= \frac{11(2\sqrt{3}+1)}{(2\sqrt{3})^2-(1^2)}=\frac{11(2\sqrt{3}+1)}{11} =2\sqrt{3}+1$

d) $\frac{2-\sqrt{2} }{3+\sqrt{2} }.\frac{3-\sqrt{2} }{3-\sqrt{2} }= \frac{(2-\sqrt{2})(3-\sqrt{2}) }{9-2} =\frac{(2-\sqrt{2})(3-\sqrt{2}) }{7}$

e) $\frac{2-2\sqrt{2} }{2-\sqrt{2} }.\frac{2+\sqrt{2}} {2+\sqrt{2} }= \frac{(2-2\sqrt{2})(2+\sqrt{2}) }{4-2} = \frac{(2-2\sqrt{2})(2+\sqrt{2}) }{2} =\frac{4 +2\sqrt{2}-4\sqrt{2}-4 }{2}=\frac{-2\sqrt{2}}{2}=-\sqrt{2}$

f)$\frac{\sqrt{3} -\sqrt{2} }{\sqrt{3}+\sqrt{2}}.\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{(\sqrt{3}-\sqrt{2})^2 }{3-2}=\frac{3-2}{3-2}=\frac{1}{1}=1$