Bài tập toán lớp 6 (Bài 6-THCS Lý Thường Kiệt)

 Đề:

Tính:

$A=\frac{3^2}{5.14}+\frac{3^2}{7.18}+\frac{3^2}{9.22}+\frac{3^2}{11.26}+\frac{3^2}{13.30}$

Bài giải:

Ta có:

$2\frac{3^2}{5.14} = \frac{3^2}{5.7}=\frac{3^2}{2}(\frac{1}{5}-\frac{1}{7})$

$2\frac{3^2}{7.18} = \frac{3^2}{7.9}=\frac{3^2}{2}(\frac{1}{7}-\frac{1}{9})$

$2\frac{3^2}{9.22} = \frac{3^2}{9.11}=\frac{3^2}{2}(\frac{1}{9}-\frac{1}{11})$

$2\frac{3^2}{11.26} = \frac{3^2}{11.13}=\frac{3^2}{2}(\frac{1}{11}-\frac{1}{13})$

$2\frac{3^2}{13.30} = \frac{3^2}{13.15}=\frac{3^2}{2}(\frac{1}{13}-\frac{1}{15})$

Cộng vế theo vế các đẳng thức trên ta có:

$\require{cancel}2A=\frac{3^2}{2}(\frac{1}{5}-\frac{1}{7})+\frac{3^2}{2}(\frac{1}{7}-\frac{1}{9})+\frac{3^2}{2}(\frac{1}{9}-\frac{1}{11})+\frac{3^2}{2}(\frac{1}{11}-\frac{1}{13})+\frac{3^2}{2}(\frac{1}{13}-\frac{1}{15})$

$=\frac{3^2}{2}(\frac{1}{5}-\cancel{\frac{1}{7}}+\cancel{\frac{1}{7}}-\cancel{\frac{1}{9}}+\cancel{\frac{1}{9}}-\cancel{\frac{1}{11}}+\cancel{\frac{1}{11}}-\cancel{\frac{1}{13}}+\cancel{\frac{1}{13}}-\frac{1}{15})$

$=\frac{3^2}{2}(\frac{1}{5}-\frac{1}{15})$

$=\frac{3^2}{2}(\frac{3}{15}-\frac{1}{15})$

$=\frac{3^2}{2}\frac{3-1}{15}$

$=\frac{3^2}{2}\frac{2}{15}$

$=\frac{3}{5}$

Vậy $A=\frac{3}{10}$

Bài viết liên quan