Question

3 intregi si 1 supra 9 -5 supra 18 - 1 intreg si 5 supra 24

Answer 1

Rezolvarea este in fotografie.

*Am gresit ceva in fotografie acolo dupa amplificare trebuia taiata partea aia si trebuie scris asa:

224-20-87/72

P.S. tot acelasi rezultat, doar ca prima data am amplficat gresit si nu am sters decat sus, fara rezultate.

Answer 2

$3\dfrac{1}{9}-\dfrac{5}{18}-1\dfrac{5}{24} = \\ \\ = 3+\dfrac{1}{9}-\dfrac{5}{18}-1-\dfrac{5}{24} = \\ \\ = 2+\dfrac{1}{9}-\dfrac{5}{18}-\dfrac{5}{24} = \\ \\ =2\dfrac{1}{9} - \dfrac{5}{18}-\dfrac{5}{24} = \\ \\ = \dfrac{2\cdot 9+1}{9}-\dfrac{5}{18}-\dfrac{5}{24} = \\ \\ = \dfrac{19}{9}-\dfrac{5}{18}-\dfrac{5}{24} \overset{(*)}{=}$

$9 = 3^2 \\ 18 = 2\cdot 3^2 \\ 24 = 2^3\cdot3\\ \\ cmmmc (9,18,24) = 2^3\cdot 3^2 = 8\cdot 9 = 72 \\ \\ \overset{(*)}{=} ^\big{2^3\slash}\dfrac{19}{3^2}-{^{^{^\big{2^2\slash}}}\dfrac{5}{2\cdot 3^2}} -{^{^{^\big{3\slash}}}\dfrac{5}{2^3\cdot 3} = = \\ \\ = \dfrac{8\cdot 19}{72}-\dfrac{4\cdot 5}{72}-\dfrac{3\cdot 5}{72} = \\ \\ = \dfrac{152-20-15}{72} = \\ \\ = \dfrac{117}{72}^{\backslash9} = \\ \\ = \dfrac{13}{8}$