Question

1/ Tính :

a) A = \(\dfrac{3737\times43-4343\times37}{2+4+6+........+2018}\)

b) B = \(\dfrac{101+100+99+..........+3+2+1}{101-100+99-98+........+3-2+1}\)

P/s : Các bạn giải hộ mình với, trong tối nay nhé !

Answer 1

a) \(A=\dfrac{3737.43-4343.37}{2+4+6+...+2018}\)

\(\Leftrightarrow A=\dfrac{3737.43-43.101.37}{2+4+6+...+2018}\)

\(\Leftrightarrow A=\dfrac{3737.43-43.3737}{2+4+6+...+2018}\)

\(\Leftrightarrow A=\dfrac{0}{2+4+6+...+2018}\)

\(\Leftrightarrow A=0\)

b) \(B=\dfrac{101+100+99+...+3+2+1}{101-100+99-98+...+3-2+1}\)

\(\Leftrightarrow B=\dfrac{\left(101+1\right).101:2}{\left(101+99+97+...+1\right)-\left(100+98+96+...+2\right)}\)

\(\Leftrightarrow B=\dfrac{5151}{\left[\left(101+1\right).51:2\right]-\left[\left(100+2\right).50:2\right]}\)

\(\Leftrightarrow B=\dfrac{5151}{2601-2550}\)

\(\Leftrightarrow B=\dfrac{5151}{51}\)

\(\Leftrightarrow B=101\)