Bài 1.29 trang 19 SBT Toán 7 Kết nối tri thức: Tính (A = left[ {left( {dfrac{1}{{81}} - dfrac{3}{{162}}} right).dfrac{{81}}{...

Tính $A = \left[ {\left( {\dfrac{1}{{81}} - \dfrac{3}{{162}}} \right).\dfrac{{81}}{{17}} + \dfrac{{35}}{{34}}} \right]:\left[ {\left( {\dfrac{9}{{51}} - \dfrac{7}{{102}}} \right).\dfrac{{102}}{5} + 2017} \right]$

- Rút gọn từng biểu thức trong ngoặc trước.

-Quy đồng mẫu số các phân số

-Thực hiện thứ tự phép tính: Nhân chia trước, cộng trừ sau.

$\begin{array}{l}\left( {\dfrac{1}{{81}} - \dfrac{3}{{162}}} \right).\dfrac{{81}}{{17}} + \dfrac{{35}}{{34}} = \left( {\dfrac{2}{{162}} - \dfrac{3}{{162}}} \right).\dfrac{{81}}{{17}} + \dfrac{{35}}{{34}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{ - 1}}{{162}}.\dfrac{{81}}{{17}} + \dfrac{{35}}{{34}} = \dfrac{{ - 1}}{{34}} + \dfrac{{35}}{{34}} = 1\\\left( {\dfrac{9}{{51}} + \dfrac{7}{{102}}} \right).\dfrac{{102}}{5} + 2017 = \left( {\dfrac{{18}}{{102}} + \dfrac{7}{{102}}} \right).\dfrac{{102}}{5} + 2017\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{25}}{{102}}.\dfrac{{102}}{5} + 2017 = 5 + 2017 = 2022\end{array}$

Vậy $A = \dfrac{1}{{2022}}$.