Question

Ma poate ajuta cineva cu aceasta matematica tare greu mai e

Answer 1

 

$\displaystyle\bf\\Folosim~formulele:\\\\\sqrt{b}=b^{\frac{1}{2}}\\\\\sqrt[\b k]{b}=b^{\frac{1}{k}}\\\\log_ab^c=c\cdot log_ab\\\\log_{a^{c}}b=\frac{1}{c}log_ab\\\\log_a^2b=\Big(log_ab\Big)^2$

.

$\displaystyle\bf\\Rezolvare:\\\\2log_5\sqrt[4]{5}+\frac{1}{2}log_{\sqrt{5}}25-log_5^2\sqrt{5}-2=\\\\=2log_55^\frac{1}{4} +\frac{1}{2}log_{5^\frac{1}{2}}25-\Big(log_55^{\frac{1}{2}}\Big)^2-2=\\\\=\frac{1}{4}\cdot2log_55 +\frac{1}{\dfrac{1}{2} }\cdot\frac{1}{2}log_{5}25-\Big(\frac{1}{2}\cdot log_55\Big)^2-2=\\\\=\frac{2}{4}\cdot1 +2\cdot\frac{1}{2}\cdot2-\Big(\frac{1}{2}\cdot1\Big)^2-2=\\\\=\frac{2}{4} +\frac{2}{2}\cdot2-\frac{1}{4}-2=\\\\=\frac{2}{4} +2-\frac{1}{4}-2=\boxed{\bf\frac{1}{4}}$