Question

Answer 1

\(\left\{{}\begin{matrix}A+B=1639\left(1\right)\\B+C=1528\left(2\right)\\C+A=1115\left(3\right)\end{matrix}\right.\\ \Rightarrow\left(1\right)+\left(2\right)+\left(3\right)=2\left(A+B+C\right)=4282\\ \Rightarrow A+B+C=2141\left(4\right)\\ \Rightarrow\left(4\right)-\left(2\right)=A+B+C-B-C=2141-1528\\ \Rightarrow A=613\)

Vậy $A=613$

Answer 2

 

⎨⎪⎩A+B=1639(1)B+C=1528(2)C+A=1115(3)⇒(1)+(2)+(3)=2(A+B+C)=4282⇒A+B+C=2141(4)⇒(4)−(2)=A+B+C−B−C=2141−1528⇒A=613{A+B=1639(1)B+C=1528(2)C+A=1115(3)⇒(1)+(2)+(3)=2(A+B+C)=4282⇒A+B+C=2141(4)⇒(4)−(2)=A+B+C−B−C=2141−1528⇒A=613

Vậy $A=613$

A+B=1639(1)