Question

Va rogggggg u urgent

Answer 1

   
[tex]\displaystyle \\ u) \\ 12 \frac{1}{3} :\left\{ \left[2 \frac{3}{4} : \left(1 \frac{2}{3} +1 \frac{7}{8}\cdot x \right)\right] \cdot \frac{8}{11} +1 \frac{2}{3} \right\}=5 \\ \\ \texttt{Introducem intregii in fractie. } \\ \\ \frac{37}{3} :\left\{ \left[ \frac{11}{4} : \left(\frac{5}{3} + \frac{15}{8}\cdot x \right)\right] \cdot \frac{8}{11} + \frac{5}{3} \right\}=5[/tex]


[tex]\displaystyle \\ \left\{ \left[ \frac{11}{4} : \left(\frac{5}{3} + \frac{15}{8}\cdot x \right)\right] \cdot \frac{8}{11} + \frac{5}{3} \right\}=\frac{37}{3} :5 \\ \\ \left[ \frac{11}{4} : \left(\frac{5}{3} + \frac{15}{8}\cdot x \right)\right] \cdot \frac{8}{11} + \frac{5}{3} =\frac{37}{15} \\ \\ \left[ \frac{11}{4} : \left(\frac{5}{3} + \frac{15}{8}\cdot x \right)\right] \cdot \frac{8}{11}=\frac{37}{15} - \frac{5}{3} [/tex]


[tex]\displaystyle \\ \left[ \frac{11}{4} : \left(\frac{5}{3} + \frac{15}{8}\cdot x \right)\right] \cdot \frac{8}{11}=\frac{37}{15} - \frac{25}{15} \\ \\ \left[ \frac{11}{4} : \left(\frac{5}{3} + \frac{15}{8}\cdot x \right)\right] \cdot \frac{8}{11}=\frac{12}{15} \\ \\ \left[ \frac{11}{4}:\left(\frac{5}{3} + \frac{15}{8}\cdot x \right)\right] =\frac{12}{15} : \frac{8}{11}\\ \\ \left[ \frac{11}{4} : \left(\frac{5}{3} + \frac{15}{8}\cdot x \right)\right] =\frac{12}{15} \cdot \frac{11}{8} [/tex]


[tex]\displaystyle \\ \frac{11}{4} : \left(\frac{5}{3} + \frac{15}{8}\cdot x \right) =\frac{33}{30} \\ \\ \frac{11}{4} : \left(\frac{5}{3} + \frac{15}{8}\cdot x \right) =\frac{11}{10} \\ \\ \left(\frac{5}{3} + \frac{15}{8}\cdot x \right) = \frac{11}{4} : \frac{11}{10} \\ \\ \frac{5}{3} + \frac{15}{8}\cdot x = \frac{11}{4} \cdot \frac{10}{11} \\ \\ \frac{5}{3} + \frac{15}{8}\cdot x = \frac{10}{4}\\ \\ \frac{5}{3} + \frac{15}{8}\cdot x = \frac{5}{2}[/tex]


[tex]\displaystyle \\ \frac{15}{8}\cdot x = \frac{5}{2} - \frac{5}{3} \\ \\ \frac{15}{8}\cdot x = \frac{15}{6} - \frac{10}{6} \\ \\ \frac{15}{8}\cdot x = \frac{5}{6} \\ \\ x= \frac{5}{6} : \frac{15}{8}=\frac{5}{6} \cdot \frac{8}{15}= \frac{1\cdot 4}{3 \cdot 3}= \boxed{\frac{4}{9}} [/tex]