Behaviour being used: Adaptive mode
Minimum fraction: 0
Question summary: Welche Äquivalenzumformungen sind richtig: {[x over 3 minus fraction numerator x minus 1 over denominator 6 end fraction equals fraction numerator x plus 1 over denominator negative 2 end fraction left right double arrow fraction numerator 2 x minus left parenthesis x minus 1 right parenthesis over denominator 6 end fraction equals fraction numerator negative 3 left parenthesis x plus 1 right parenthesis over denominator 6 end fraction]; [x over 3 minus fraction numerator x minus 1 over denominator 6 end fraction equals fraction numerator x plus 1 over denominator negative 2 end fraction left right double arrow fraction numerator 2 x minus left parenthesis x minus 1 right parenthesis over denominator 6 end fraction equals fraction numerator 3 left parenthesis x plus 1 right parenthesis over denominator 6 end fraction]; [x over 3 minus fraction numerator x minus 1 over denominator 6 end fraction equals fraction numerator x plus 1 over denominator negative 2 end fraction left right double arrow fraction numerator 2 x minus x plus 1 over denominator 6 end fraction equals fraction numerator 3 x plus 3 over denominator 6 end fraction]} -> {falsch; richtig}
Right answer summary: [x over 3 minus fraction numerator x minus 1 over denominator 6 end fraction equals fraction numerator x plus 1 over denominator negative 2 end fraction left right double arrow fraction numerator 2 x minus left parenthesis x minus 1 right parenthesis over denominator 6 end fraction equals fraction numerator negative 3 left parenthesis x plus 1 right parenthesis over denominator 6 end fraction] -> richtig; [x over 3 minus fraction numerator x minus 1 over denominator 6 end fraction equals fraction numerator x plus 1 over denominator negative 2 end fraction left right double arrow fraction numerator 2 x minus left parenthesis x minus 1 right parenthesis over denominator 6 end fraction equals fraction numerator 3 left parenthesis x plus 1 right parenthesis over denominator 6 end fraction] -> falsch; [x over 3 minus fraction numerator x minus 1 over denominator 6 end fraction equals fraction numerator x plus 1 over denominator negative 2 end fraction left right double arrow fraction numerator 2 x minus x plus 1 over denominator 6 end fraction equals fraction numerator 3 x plus 3 over denominator 6 end fraction] -> falsch
Question state: todo