Giải:
Ta có:
\(\left(a+b+c+d\right)-\left(a+c+d\right)._{\left(1\right)}\)
\(=a+b+c+d-a-c-d.\)
\(=\left(a-a\right)+\left(c-c\right)+\left(d-d\right)+b.\)
\(=0+0+0+b=b.\)
Thay số vào \(_{\left(1\right)}\)\(\Rightarrow1-2=b\Rightarrow b=-1\in Z.\)
\(\left(a+b+c+d\right)-\left(a+b+d\right)._{\left(2\right)}\)
\(=a+b+c+d-a-b-d.\)
\(=\left(a-a\right)+\left(b-b\right)+\left(d+d\right)+c.\)
\(=0+0+0+c=c.\)
Thay số vào \(_{\left(2\right)}\)\(\Rightarrow1-3=c\Rightarrow c=-2\in Z.\)
\(\left(a+b+c+d\right)-\left(a+b+c\right)_{\left(3\right)}.\)
\(=a+b+c+d-a-b-c.\)
\(=\left(a-a\right)+\left(b-b\right)+\left(c-c\right)+d.\)
\(=0+0+0+d=d.\)
Thay số vào \(_{\left(3\right)}\)\(\Rightarrow1-4=d\Rightarrow d=-3\in Z.\)
\(\Rightarrow a+b+c+d=1.\)
\(a+\left(-1\right)+\left(-2\right)+\left(-3\right)=1.\)
\(\Rightarrow a=1-\left(-1\right)-\left(-2\right)-\left(-3\right).\)
\(\Rightarrow a=1+1+2+3=7\in Z.\)
Vậy \(\left\{a;b;c;d\right\}=\left\{7;-1;-2;-3\right\}.\)
Do a + b + c + d = 1 mà a + c + d = 2
=> b = 1 - 2 = -1
=> c = 1 - 3 = -2
=> d = 1 - 4 = -3
=> a = 1 - (-1 - 2 - 3) = 7
@Valentine
