Question

\(\dfrac{\left(1+2+...+100\right).\left(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}\)

thực hiện phép tính

Answer 1

\(\dfrac{\left(1+2+...+100\right)\left(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(6,3.12-21.3,6\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}\)

\(=\dfrac{\left(1+2+...+100\right)\left(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}\right).0}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}\)

\(=\dfrac{0}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}}=0\)

Answer 2

Ta xét :

\(\left(6,3.12-21.3,6\right)=75,6-75,6=0\)

Từ đây ta thấy các tích nhân với 0 sẽ bằng 0 mà 0 chia cho số nào cũng vẫn bằng 0

\(\Rightarrow\) phép tính đó bằng 0

Vậy............

Answer 3

\(\dfrac{\left(1+2+.....+100\right).\left(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(6,3.12-21.3,6\right)}{\dfrac{1}{2}+\dfrac{1}{3}+......+\dfrac{1}{100}}\)\(=\dfrac{\left(1+2+3+.....+100\right).\left(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}\right).0}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{100}}\)

\(=\dfrac{A.0}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{100}}\)

\(=\dfrac{0}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{100}}=0\)