2015-08-17から1日間の記事一覧
前回の qfq では,例えば (%i1) f:abs(x-a)=abs(x-b)+c; (%o1) abs(x - a) = abs(x - b) + c (%i2) qfq(f); (%o2) ((0 < c) %and (a = b - c) %and (x = b)) %or ((0 < c) %and (b = a - c) %and (x = a - c)) %or ((a = b) %and (c = 0) %and (x = b)) %or …