Question

va rog cat mai repede si daca se poate un raspuns clar<br />ex 5 si 6

Answer 1

$5)n=\big(\frac{5\sqrt3}{\sqrt8}+\frac{3\sqrt3}{5\sqrt2}-\frac{7\sqrt3}{\sqrt2}\big):\sqrt3=\big(\frac{5\sqrt3}{\sqrt8}+\frac{3\sqrt3}{5\sqrt2}-\frac{7\sqrt3}{\sqrt2}\big)\cdot\frac{1}{\sqrt3}=\frac{5\sqrt3}{\sqrt8}\cdot\frac{1}{\sqrt3}\\\\+\frac{3\sqrt3}{5\sqrt2}\cdot\frac{1}{\sqrt3}-\frac{7\sqrt3}{\sqrt2}\cdot\frac{1}{\sqrt3}=\frac{5}{\sqrt8}+\frac{3}{5\sqrt2}-\frac{7}{\sqrt2}$

$n\sqrt2=\big(\frac{5}{\sqrt8}+\frac{3}{5\sqrt2}-\frac{7}{\sqrt2}\big)\cdot\sqrt2=\frac{5}{\sqrt8}\cdot\sqrt2+\frac{3}{5\sqrt2}\cdot\sqrt2-\frac{7}{\sqrt2}\cdot\sqrt2\\\\=\frac{5}{\sqrt4}+\frac{3}{5}-7=\frac{5\cdot 5+3\cdot2-7\cdot10}{10}=\frac{-39}{10}$

$-40<-39<-10\text{ }|:10\\ -4<\frac{-39}{10}<-1\Leftrightarrow [n\sqrt2]=-4$

$6) x-\frac{1}{x}=3\\\\ (x-\frac{1}{x})^2=3^2\\\\ x^2-2\cdot x\cdot\frac{1}{x}+\frac{1}{x^2}=9\\\\ x^2+\frac{1}{x^2}-2=9\\\\ x^2+\frac{1}{x^2}=9+2=11\\\\ p\%\text{ din }300=11\Leftrightarrow \frac{p}{100}\cdot 300=11\Leftrightarrow p\cdot3=11\Leftrightarrow p=\frac{11}{3}=3(6)$