Question

Ma poate ajuta si pe mine cineva cu aceste doua exercitii? Va rog !

Answer 1

 

$\displaystyle\\1)\\a)\\\\\sqrt{18}-\sqrt{32}-\sqrt{50}+\sqrt{72}=\\\\   =\sqrt{9\cdot2}-\sqrt{16\cdot2}-\sqrt{25\cdot2}+\sqrt{36\cdot2}=\\\\=3\sqrt{2}-4\sqrt{2}-5\sqrt{2}+6\sqrt{2}=\\\\=3\sqrt{2}+6\sqrt{2}-4\sqrt{2}-5\sqrt{2}=9\sqrt{2}-9\sqrt{2}=0\sqrt{2}=\boxed{0}$

$\displaystyle\\b)\\\\2\sqrt{14}-\sqrt{\frac{7}{2}}-5\sqrt{\frac{8}{7}}=\\\\=2\sqrt{14}-\frac{\sqrt{7}}{\sqrt{2}}-\frac{5\sqrt{8}}{\sqrt{7}}=\\\\=2\sqrt{14}-\frac{\sqrt{7}\cdot\sqrt{2}}{2}-\frac{5\sqrt{4\cdot2}~\cdot\sqrt{7}}{7}=\\\\=2\sqrt{14}-\frac{\sqrt{14}}{2}-\frac{5\cdot2\sqrt{14}}{7}=\\\\=\frac{14\cdot2\sqrt{14}}{14}-\frac{7\sqrt{14}}{14}-\frac{2\cdot5\cdot2\sqrt{14}}{14}=\\\\=\frac{28\sqrt{14}-7\sqrt{14}-20\sqrt{14}}{14}= \boxed{\frac{\sqrt{14}}{14}}$

$\displaystyle\\c)\\\\8^{^\frac{\b1}{\b3}}-\Big(\frac{1}{2}\Big)^{\b-\b1}=\sqrt[\b3]{8}-\Big(\frac{2}{1}\Big)^{\b1}=2-2=\boxed{0}\\\\\\d)\\\\\frac{\sqrt{50}+\sqrt{18}}{\sqrt{32}-\sqrt{8}}=\frac{5\sqrt{2}+3\sqrt{2}}{4\sqrt{2}-2\sqrt{2}}=\frac{\sqrt{2}(5+3)}{\sqrt{2}(4-2)}=\frac{8}{2}=\boxed{4}$

$\displaystyle\\\\2)\\\\a)\\\\\log_{11}11+\log_7\frac{1}{7}=\log_{11}11^1+\log_7 7^{-1}=1-1=\boxed{0}\\\\ b)\\\\\log_212+\log_214-\log_221=\log_2\frac{12\cdot14}{21}=\\\\=\log_2\frac{(3\cdot4)\cdot(2\cdot7)}{3\cdot7}=\log_2(4\cdot 2)=\log_2(8)=\boxed{3}\\\\ c)\\\\\log_381-\log_525-\lg100000=4+2-5=\boxed{1}$