Bài tập toán lớp 6 (Bài 6-THCS Ngô Sĩ Liên)

 Đề bài:

Tính hợp lý: $\frac{1}{2.3}-\frac{8}{3.5}+\frac{4}{5.9}-\frac{22}{9.13}+\frac{31}{13.18}$

Bài giải:

Ta có:

$\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}$

$\frac{8}{3.5}=\frac{1}{3}+\frac{1}{5}$

$\frac{4}{5.9}=\frac{1}{5}-\frac{1}{9}$

$\frac{22}{9.13}=\frac{1}{9}+\frac{1}{13}$

$\frac{31}{13.18}=\frac{1}{13}+\frac{1}{18}$

Suy ra:

$\frac{1}{2.3}-\frac{8}{3.5}+\frac{4}{5.9}-\frac{22}{9.13}+\frac{31}{13.18}$

$=(\frac{1}{2}-\frac{1}{3})-(\frac{1}{3}+\frac{1}{5})+(\frac{1}{5}-\frac{1}{9})-(\frac{1}{9}+\frac{1}{13})+(\frac{1}{13}+\frac{1}{18})$

$=\frac{1}{2}-\frac{1}{3}-\frac{1}{3}-\cancel{\frac{1}{5}}+\cancel{\frac{1}{5}}-\frac{1}{9}-\frac{1}{9}-\cancel{\frac{1}{13}}+\cancel{\frac{1}{13}}+\frac{1}{18}$

$=\frac{1}{2}-2\frac{1}{3}-2\frac{1}{9}+\frac{1}{18}$

$=\frac{9}{18}-2\frac{6}{18}-2\frac{2}{18}+\frac{1}{18}$

$=\frac{9-12-4+1}{18}$

$=\frac{-6}{18}$

$=\frac{-1}{3}$