Question

repede....rezolvare completa.....dau coroana​

Answer 1

Răspuns:

Explicație pas cu pas:

$=(\frac{\sqrt{4}}{\sqrt{12}}-\frac{\sqrt{3}}{\sqrt{12}}+\frac{\sqrt{5}}{\sqrt{20}}-\frac{\sqrt{4} }{\sqrt{20}} +\frac{\sqrt{6}}{30}-\frac{\sqrt{5}}{\sqrt{30}}+\frac{\sqrt{7}}{\sqrt{42}}-\frac{\sqrt{6}}{\sqrt{42}}+\frac{\sqrt{8}}{\sqrt{56}}-\frac{\sqrt{7}}{\sqrt{56}}+\frac{\sqrt{9}}{\sqrt{72}}-\frac{\sqrt{8}}{\sqrt{72}}):\frac{2}{\sqrt{27}+3 }=~(\frac{1}{\sqrt{3}}- \frac{1}{\sqrt{4}}+\frac{1}{\sqrt{4}}-\frac{1}{\sqrt{5}}+ \frac{1}{\sqrt{5}}-\frac{1}{\sqrt{6}}+ \frac{1}{\sqrt{6}}-\frac{1}{\sqrt{7}}+$

$+\frac{1}{\sqrt{7}}-\frac{1}{\sqrt{8}}+\frac{1}{\sqrt{8}}-\frac{1}{\sqrt{9}})*\frac{\sqrt{27}+3 }{2}=( \frac{1}{\sqrt{3}}-\frac{1}{\sqrt{9}})*\frac{3*(\sqrt{3}+1 )}{2}= \frac{\sqrt{3} -1}{\sqrt{9} } *\frac{3*(\sqrt{3}+1 )}{2}=\frac{(\sqrt{3} -1)(\sqrt{3}+1) }{2}=\frac{(\sqrt{3})^{2}-1^{2} }{2}=\frac{3-1}{2}=1$