Question

BT: Rút gọn: \(A=\dfrac{\left(1+2+3+...+99+100\right)\times\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{2}\right)\times\left(63\times1.2-21\times3,6+1\right)}{1-2+3-4+5-6+...+99-100}\)

Giúp mình với mình cần gấp!!! Tối mai mình học rồi!!! Cảm ơn các bạn nhiều!!!

Answer 1

101/12

Answer 2

bang 0

Answer 3

Mình cần cách trình bày!

Answer 4

\(A=\dfrac{\left(1+2+3+...+99+100\right)\cdot\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{2}\right)\cdot\left(63\cdot1,2-21\cdot3,6+1\right)}{1-2+3-4+5-6+...+99-100}\)

Gọi \(\left(1+2+3+...+99+100\right)\cdot\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{2}\right)\cdot\left(63\cdot1,2-21\cdot3,6+1\right)\) là \(M\)

Ta có:

\(M=\left(1+2+3+...+99+100\right)\cdot\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{2}\right)\cdot\left(63\cdot1,2-21\cdot3,6+1\right)\\ =\left(\dfrac{100\cdot101}{2}\right)\cdot\left(\dfrac{3}{12}+\dfrac{2}{12}-\dfrac{6}{12}\right)\cdot\left(63\cdot1,2-21\cdot3\cdot1,2+1\right)\\ =5050\cdot\dfrac{-1}{12}\cdot\left(63\cdot1,2-63\cdot1,2+1\right)\\ =\dfrac{-2525}{6}\cdot\left(0+1\right)\\ =\dfrac{-2525}{6}\cdot1\\ =\dfrac{-2525}{6}\)\(A=\dfrac{\dfrac{-2525}{6}}{1-2+3-4+5-6+...+99-100}\\ =\dfrac{\dfrac{-2525}{6}}{\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)}\\ =\dfrac{\dfrac{-2525}{6}}{-50}=\dfrac{101}{12}\)