[tex]\it 1.\\ \\ \dfrac{1}{24}:\Big(-\dfrac{1}{8}\Big)+\dfrac{1}{15}\cdot\dfrac{5}{2}=-\dfrac{1}{24}\cdot\dfrac{8}{1}+\dfrac{5}{30}=-\dfrac{\ 8^{(4}}{24}+\dfrac{\ 5^{(5}}{30}=-\dfrac{2}{6}+\dfrac{1}{6}=\\ \\ \\ =\dfrac{-2+1}{6}=-\dfrac{1}{6}[/tex]
[tex]\it 4.\\ \\ Not\breve am \ \ x\ -\ pre\c{\it t}ul\ ini\c{\it t}ial\ .\\ \\ I)\ Dup\breve a\ scumpire,\ \ pre \c{\it t}ul\ devine: \\ \\ x+\dfrac{\ \ 10^{(10}}{100}x\ =\ \ ^{10)}x+\dfrac{1}{10}x=\dfrac{11}{10}x[/tex]
[tex]\it\ \ II)\ \ Dup\breve a\ ieftinire,\ pre\c{\it t}ul\ devine:\\ \\ \dfrac{11}{10}x-\dfrac{1\not0}{100}\cdot\dfrac{11}{1\not0}x=\dfrac{^{10)}11}{\ \ 10}x-\dfrac{11}{100}x=\dfrac{99}{100}x[/tex]
Dar, ultimul preț este 99 lei, deci:
[tex]\it \dfrac{99}{100}x=99 \Rightarrow x=99\cdot\dfrac{100}{99} \Rightarrow x=100\ \ell ei[/tex]
[tex]\it 7.\\ \\ \mathcal{A}=\dfrac{d_1\cdot d_2}{2} \Rightarrow 12=\dfrac{6\cdot d_2}{2} \Rightarrow 12=3\cdot d_2 \ \Big|_{:3} \Rightarrow d_2=4\ cm[/tex]
