Các câu hỏi tương tự
Tính giá trị sau một cách hợp lí
M = \(\frac{\frac{1}{3}-\frac{1}{7}+\frac{1}{13}}{\frac{7}{3}-\frac{7}{7}+\frac{7}{13}}+\frac{\frac{2}{11}+\frac{4}{17}-\frac{6}{19}}{\frac{7}{11}-\frac{21}{19}+\frac{14}{17}}\)-19\(\frac{3}{7}\)
N = \(\frac{49.\left(-113\right)+98.37-147}{7^3.\left(-9\right)}\)
\(\frac{4}{7}.\frac{3}{5}.\frac{7}{4}.\left(-20\right).\frac{-11}{12}\)Tính nhazz
\(\frac{7}{13}.\frac{5}{19}+\frac{7}{19}.\frac{8}{13}-\frac{37}{19}\)
\(\frac{7}{13}.\frac{5}{19}+\frac{7}{19}.\frac{8}{13}-3\frac{7}{19}\)
tính giùm mình,cảm ơn nhé
Tính hợp lí: \(\frac{12}{9}.\frac{7}{15}.\frac{-13}{17}.\frac{19}{12}.\frac{17}{13}\)
Tính hợp lý:a) A frac{101}{19}. frac{61}{218}- frac{101}{218}. frac{42}{19}+ frac{117}{218}b) B left(frac{5}{2011^2}+frac{7}{2012^2}-frac{9}{2013^2}right). left(frac{4}{5}-frac{3}{4}-frac{1}{20}right)
Đọc tiếp
Tính hợp lý:
a) A= \(\frac{101}{19}\). \(\frac{61}{218}\)\(-\) \(\frac{101}{218}\). \(\frac{42}{19}\)+ \(\frac{117}{218}\)
b) B= \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right)\). \(\left(\frac{4}{5}-\frac{3}{4}-\frac{1}{20}\right)\)
Tính bằng cách hợp lý:
a, \(\frac{4}{19}.\frac{-3}{7}+\frac{-3}{7}.\frac{15}{19}+\frac{5}{7}\)
b, \(\frac{-3}{5}.\frac{5}{7}+\frac{-3}{5}.\frac{3}{7}+\frac{-3}{5}.\frac{6}{7}\)
( Các bn nhớ tính bằng cách hợp lý đó nha)
\(B=\frac{8}{3}.\frac{2}{5}.\frac{3}{8}.10.\frac{19}{92}\)
\(C=\frac{-5}{7}.\frac{2}{7}+\frac{-5}{7}.\frac{9}{14}+1\frac{5}{7}\)
\(D=\frac{12}{19}.\frac{7}{15}.\frac{-13}{17}.\frac{19}{12}.\frac{17}{13}\)
\(E=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
Tính bằng cách thuận tiện
1,\(\frac{13}{19}\left(\frac{19}{26}-\frac{71}{43}\right)+\frac{71}{43}\left(\frac{13}{19}-\frac{86}{71}\right)\)
2,\(\frac{5}{7}\left(\frac{1}{2}-\frac{1}{3}+\frac{4}{7}\right)+\left(\frac{1}{2}-\frac{1}{3}-\frac{4}{7}\right):\frac{7}{5}\)
a)Tính\(\frac{\left(17\frac{8}{19}-16\frac{9}{18}\right)\left(17,5+16\frac{17}{51}-32\frac{15}{22}\right)}{\frac{7}{3.13}+\frac{7}{13.23}+\frac{7}{23.33}}\)
b) Chứng tò rằng:\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)