Đơn giản như đang giỡn
Ta có: \(A=\dfrac{101+100+99+...+3+2+1}{101-100+99-98+...+3-2+1}\)
\(=\dfrac{\dfrac{101.102}{2}}{\left(101-100\right)+\left(99-98\right)+...+\left(3-2\right)+1}\)
\(=\dfrac{101.51}{51}=101\)
Vậy \(A=101\)
\(\dfrac{101+100+99+...+3+2+1}{101-100+99-98+...+3-2+1}\)
\(=\dfrac{\left(101+1\right).101:2}{1+1+1+...+1}\)
\(=\dfrac{5151}{51}=101\)