E.469. Determinați toate valorile posibile ale numerelor aaa și bbb pentru care: 1+{2×[3+(a×b−4)×5]:6}×7=78.1+\{2 \times [3+(a \times b - 4) \times 5]:6\} \times 7 = 78.1+{2×[3+(a×b−4)×5]:6}×7=78.
a×b=10⇒(a;b)∈{(1;10),(10;1),(2;5),(5;2)}.a \times b = 10 \Rightarrow (a;b) \in \{(1;10), (10;1), (2;5), (5;2)\}.a×b=10⇒(a;b)∈{(1;10),(10;1),(2;5),(5;2)}.
{2×[3+(a×b−4)×5]:6}×7=77.\{2 \times [3+(a \times b - 4) \times 5]:6\} \times 7 = 77.{2×[3+(a×b−4)×5]:6}×7=77. 2×[3+(a×b−4)×5]:6=11.2 \times [3+(a \times b - 4) \times 5]:6 = 11.2×[3+(a×b−4)×5]:6=11. 3+(a×b−4)×5=33.3+(a \times b - 4) \times 5 = 33.3+(a×b−4)×5=33. (a×b−4)×5=30.(a \times b - 4) \times 5 = 30.(a×b−4)×5=30. a×b−4=6⇒a×b=10⇒(a;b)∈{(1;10),(10;1),(2;5),(5;2)}.a \times b - 4 =6 \Rightarrow \boxed{a \times b = 10} \Rightarrow \boxed{(a;b) \in \{(1;10), (10;1), (2;5), (5;2)\}}.a×b−4=6⇒a×b=10⇒(a;b)∈{(1;10),(10;1),(2;5),(5;2)}.