[tex]\it a)\ \ 2\cdot2\dfrac{1}{7}=2\cdot\dfrac{15}{7}\\ \\ \\ 0,(7)=\dfrac{7}{9}\\ \\ \\ 2,7(2)=2\dfrac{72-7}{90}=2\dfrac{\ 65^{(5}}{90}=2\dfrac{13}{18}=\dfrac{49}{18}[/tex]
Acum, ecuația se scrie:
[tex]\it \dfrac{2\cdot\dfrac{15}{7}}{\dfrac{7}{9}}=\dfrac{x}{\dfrac{49}{18}} \Rightarrow x= 2\cdot\dfrac{15}{7}\cdot\dfrac{49}{18}\cdot\dfrac{9}{7}=15[/tex]
[tex]\it b)\ \ \dfrac{10}{15}=\dfrac{3n+5}{21} \Rightarrow 3n+5=\dfrac{10\cdot21}{15} \Rightarrow 3n+5=\dfrac{210}{15} \Rightarrow \\ \\ \\ \Rightarrow3n+5=14|_{-5} \Rightarrow3n=9|_{:3} \Rightarrow n=3[/tex]
[tex]\it |x+4|-7=-2|x+4|+8 \Rightarrow |x+4|+2|x+4|=8+7 \Rightarrow3|x+4|=15|_{:3}\Rightarrow\\ \\ \Rightarrow|x+4|=5\Rightarrow x+4=\pm5\Rightarrow x+4\in\{-5,\ 5\}|_{-4}\Rightarrow x\in\{-9,\ 1\}[/tex]
Conform enunțului, vom avea a=1, b = |-9| = 9
[tex]\it \overline{abcc}\ \vdots\ 5 \Rightarrow c=5\\ \\ Deci\ \ \overline{abcc}=1955[/tex]
Prin urmare, data cerută este 15 - 03 - 1955
