primul chenar e modelul.Va rog ajutati.ma dau coroana

[tex]4.(3) = \frac{43 - 4}{9} = \frac{ {39}^{( \div 3} }{9} = \boxed{ \frac{13}{3} } \\ 12.(3) = \frac{123 - 12}{9} = \frac{ {111}^{( \div 3} }{9} = \boxed{ \frac{37}{3} } \\ 2.(6) = \frac{26 - 2}{9} = \frac{ {24}^{( \div 3} }{9} = \boxed{ \frac{8}{3} } \\ 1.(32) = \frac{132 - 1}{99} = \boxed{ \frac{131}{99} } \\ 2.(02) = \frac{202 - 2}{99} = \boxed{\frac{200}{99} } \\ 1.2(3) = \frac{123 - 12}{90} = \frac{ {111}^{( \div 3} }{90} = \boxed{\frac{37}{30} } \\ 0.7(3) = \frac{73 - 7}{90} = \frac{ {66}^{( \div 3} }{90} = \boxed{ \frac{22}{30} } \\ 21.21(3) = \frac{21213 - 2121}{900} = \frac{ {19092}^{( \div 12} }{900} = \boxed{ \frac{1591}{75} } \\ 1.3(13) = \frac{1313 - 13}{990} = \frac{ {1300}^{( \div 10} }{990} = \boxed{ \frac{130}{99} } \\ 1.60(73) = \frac{16073 - 160}{9900} = \boxed{\frac{ {15913} }{9900} }[/tex]