Question

Tính Nhanh

\(\frac{\left(1+2+3+...+99+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)

Answer 1

Ta có: \(63.1,2-21.3,6=0,9.7.10.1,2-21.3,6\)

\(=6,3.12-21.3,6\)

\(=0,9.7.4.3-7.3.0,9.4\)

\(=6,3.12-6,3.12\)

\(=0\)

\(\Rightarrow\frac{\left(1+2+...+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)0}{1-2+3-4+...+99-100}=0\)

Vậy \(\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}=0\)

Answer 2

Dãy tính trên có (63.1,2 - 21.3,6) =0 nên kết quả là 0

Answer 3

0

Answer 4

0