Đáp án:
$ D = d -(a+c) +(b+d) -b+(b-c)$
$ = d - a -c +b +d -b +b -c$
$ = 2d -a -2c +b$
Thay $a=1\dfrac{1}{3} ; b=\dfrac{7}{2} ; c = -\dfrac{5}{6} ; d =\dfrac{1}{12}$, ta có:
$ = 2.\dfrac{1}{12} -1\dfrac{1}{3} - 2 .\dfrac{-5}{6} + \dfrac{7}{2}$
$ = \dfrac{1}{6} -\dfrac{4}{3} + \dfrac{5}{3} + \dfrac{7}{2}$
$ = \dfrac{1}{6} - \dfrac{8}{6} + \dfrac{10}{6} + \dfrac{21}{6}$
$ = 4$
Vậy $D=4$ , với $a=1\dfrac{1}{3} ; b=\dfrac{7}{2} ; c = -\dfrac{5}{6} ; d =\dfrac{1}{12}$
$E = c - (a+b-d) +a+(a-b)$
$ = c - a -b +d +a +a -b$
$ = c +a -2b +d$
Thay $a =-1\dfrac{1}{3} ; b = 3\dfrac{1}{2} ;c= -\dfrac{5}{3} ;d =\dfrac{5}{12}$,ta có :
$ = \dfrac{-5}{3} +(-1\dfrac{1}{3}) - 2 . 3\dfrac{1}{2} + \dfrac{5}{12}$
$ = -\dfrac{5}{3} + (-\dfrac{4}{3}) - 2 . \dfrac{7}{2} + \dfrac{5}{12}$
$ = -\dfrac{5}{3} +(-\dfrac{4}{3} - 7 + \dfrac{5}{12}$
$ = -\dfrac{20}{12} + (-\dfrac{16}{12}) - \dfrac{84}{12} + \dfrac{5}{12}$
$ = -\dfrac{115}{12}$
Vậy $E = -\dfrac{115}{12}$ , với $a =-1\dfrac{1}{3} ; b = 3\dfrac{1}{2} ;c= -\dfrac{5}{3} ;d =\dfrac{5}{12}$
