Giải bài 110 trang 49 - Sách giáo khoa Toán 6 tập 2

Đề bài

   Tính giá trị các biểu thức sau:

Hướng dẫn giải

      Giải: 

   \( A= 11\dfrac{3}{13}-(2\dfrac{4}{7}+5\dfrac{3}{13})=(11\dfrac{3}{13}-5\dfrac{3}{13})-2\dfrac{4}{7}=6-2\dfrac{4}{7}=5\dfrac{7}{7}-2\dfrac{4}{7}=3\dfrac{3}{7}\) \(B= (6\dfrac{4}{9}+3\dfrac{7}{11}-4\dfrac{4}{9}=(6\dfrac{4}{9}-4\dfrac{4}{9})+3\dfrac{7}{11}\)

   \(=2+3\dfrac{7}{11}=2+3+\dfrac{7}{11}=5\dfrac{7}{11}\)

 \(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}.\dfrac{9}{11}+1\dfrac{5}{7}=\dfrac{-5}{7}.(\dfrac{2}{11}+\dfrac{9}{11})+1\dfrac{5}{7}\)

   \(=\dfrac{-5}{7}.\dfrac{11}{11}+1\dfrac{5}{7}=\dfrac{-5}{7}+1+\dfrac{5}{7}\)

 \(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}=(\dfrac{7}{10}.20)(\dfrac{8}{3}.\dfrac{3}{8}).\dfrac{5}{28}\)

   \(14.1.\dfrac{5}{28}=14.1.\dfrac{5}{28}=\dfrac{5}{2}=2,5\)

\(E=(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97})(\dfrac{1}{3}-0,25-\dfrac{1}{2})=(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97})(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12})\)\(=(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}).0=0\)