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\(\dfrac{4}{7}.\dfrac{89}{5}-\dfrac{4}{5}.\dfrac{82}{7}\)
\(=\dfrac{4}{7}.\dfrac{89}{5}-\dfrac{4}{7}.\dfrac{82}{5}\)
\(=\dfrac{4}{7}.\left(\dfrac{89}{5}-\dfrac{82}{5}\right)\)
\(=\dfrac{4}{7}.\dfrac{7}{5}\)
\(=\dfrac{4}{5}\)
\(\dfrac{5}{7}.\dfrac{-4}{19}+\dfrac{-15}{7}.\dfrac{5}{19}\)
\(=\dfrac{5}{7}.\dfrac{-4}{19}+\dfrac{5}{7}.\dfrac{-15}{19}\)
\(=\dfrac{5}{7}.\left(\dfrac{-4}{19}+\dfrac{-15}{19}\right)\)
\(=\dfrac{5}{7}.\left(-1\right)\)
\(=\dfrac{-5}{7}\)
\(-2\dfrac{1}{4}.\)\(\left(3\dfrac{5}{12}-1\dfrac{2}{9}\right)\)
=\(\dfrac{-9}{4}\).\(\left(\dfrac{41}{12}-\dfrac{11}{9}\right)\)
=\(\dfrac{-9}{4}.\dfrac{41}{12}-\dfrac{-9}{4}.\dfrac{11}{9}\)
=\(\dfrac{-123}{16}-\dfrac{-11}{4}\)
=\(\dfrac{-123}{16}-\dfrac{-44}{16}\)
=\(\dfrac{-79}{16}\)
\(\left(-25\%+0,75+\dfrac{7}{12}\right)\div\left(-2\dfrac{1}{8}\right)\)
=\(\left(\dfrac{-1}{4}+\dfrac{3}{4}+\dfrac{7}{12}\right)\div\left(\dfrac{-17}{8}\right)\)
=\(\left(\dfrac{-3}{12}+\dfrac{9}{12}+\dfrac{7}{12}\right).\dfrac{-8}{17}\)
=\(\dfrac{13}{12}.\dfrac{-8}{17}=\dfrac{-26}{51}\)
a) \(6\dfrac{5}{7}-\left(1\dfrac{3}{4}+2\dfrac{5}{7}\right)\)
\(=6\dfrac{5}{7}-1\dfrac{3}{4}-2\dfrac{5}{7}\)
\(=\left(6\dfrac{5}{7}-2\dfrac{5}{7}\right)-1\dfrac{3}{4}\)
\(=4-1\dfrac{3}{4}\)
\(=3\dfrac{3}{4}\)
b) \(7\dfrac{5}{11}-\left(2\dfrac{3}{7}+3\dfrac{5}{11}\right)\)
\(=7\dfrac{5}{11}-2\dfrac{3}{7}-3\dfrac{5}{11}\)
\(=\left(7\dfrac{5}{11}-3\dfrac{5}{11}\right)-2\dfrac{3}{7}\)
\(=4-2\dfrac{3}{7}\)
\(=2\dfrac{3}{7}\)
\(M=\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}=\frac{3\left(\frac{1}{5}+\frac{1}{7}-\frac{3}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{3}{4}\) \(\frac{3}{4}\) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=2-\frac{2}{101}=\frac{200}{101}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(B=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(B=2.\frac{100}{101}=\frac{200}{101}\)
5, \(\dfrac{1}{7}.\dfrac{2}{5}+\dfrac{1}{7}.\dfrac{1}{5}+\dfrac{1}{7}.\dfrac{4}{5}\)
= \(\dfrac{1}{7}.\left(\dfrac{2}{5}+\dfrac{1}{5}+\dfrac{4}{5}\right)\)
= \(\dfrac{1}{7}\).1
=\(\dfrac{1}{7}\)
5) \(\dfrac{1}{7}.\dfrac{2}{5}+\dfrac{1}{7}.\dfrac{1}{5}+\dfrac{1}{7}.\dfrac{4}{5}\) = \(\dfrac{1}{7}.\left(\dfrac{2}{5}+\dfrac{1}{5}+\dfrac{4}{5}\right)\)
= \(\dfrac{1}{7}.\dfrac{7}{5}=\dfrac{1}{5}\)
a) -3/7 + 5/13 + -4/7 = -3/7 + 4/7 + 5/13 = -1 + 5/13 = -13/13 + 5/13 = -8/13 b) -5/21 + -2/21 + 8/24 = -7/21 + 1/3 = -1/3 + 1/3 = 0
a) \(\dfrac{-3}{7}+\dfrac{5}{13}+\dfrac{-4}{7}\)
\(=\left(\dfrac{-3}{7}+\dfrac{-4}{7}\right)+\dfrac{5}{13}\)
\(=\left(-1\right)+\dfrac{5}{13}\)
\(=\dfrac{-8}{13}\)
b) \(\dfrac{-5}{21}+\dfrac{-2}{21}+\dfrac{8}{24}\)
\(=\left(\dfrac{-5}{21}+\dfrac{-2}{21}\right)+\dfrac{1}{3}\)
\(=\dfrac{-1}{3}+\dfrac{1}{3}\)
\(=0\)
dễ lắm bạn ơi bọn mình học rùi
bn lấy tử số của từng mỗi rồi chia cho mẫu của từng cái rồi rút gọn lafd ra
\(\dfrac{\dfrac{5}{7}+\dfrac{5}{9}+\dfrac{5}{11}}{\dfrac{25}{7}+\dfrac{25}{9}+\dfrac{25}{11}}\cdot\dfrac{4+\dfrac{4}{73}-\dfrac{4}{115}}{5+\dfrac{5}{73}-\dfrac{1}{23}}\)
=\(\dfrac{5\cdot\left(\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{11}\right)}{25\left(\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{11}\right)}\cdot\dfrac{4\left(1+\dfrac{1}{73}+\dfrac{1}{115}\right)}{5+\dfrac{5}{73}-\dfrac{5}{115}}\)
=\(\dfrac{5}{25}\cdot\dfrac{4\left(1+\dfrac{1}{73}-\dfrac{1}{115}\right)}{5\left(1+\dfrac{1}{73}-\dfrac{1}{115}\right)}\)
=\(\dfrac{1}{5}\cdot\dfrac{4}{5}\)=\(\dfrac{4}{25}\)
Có:
\(\dfrac{4}{7}.\dfrac{89}{5}-\dfrac{4}{5}.\dfrac{82}{7}\)
\(=\dfrac{4}{7}.\dfrac{89}{5}-\dfrac{4.82}{5.7}=\dfrac{4}{7}.\dfrac{89}{5}-\dfrac{4.82}{7.5}=\dfrac{4}{7}.\dfrac{89}{5}-\dfrac{4}{7}.\dfrac{89}{5}\)
\(=\dfrac{4}{7}.\left(\dfrac{89}{5}-\dfrac{82}{5}\right)=\dfrac{4}{7}.\dfrac{7}{5}=\dfrac{4}{5}\)
Chúc bạn học tốt!
\(\dfrac{4}{7}\) .\(\dfrac{89}{5}\) - \(\dfrac{4}{5}\). \(\dfrac{82}{7}\)
= \(\dfrac{4.89}{7.5}\) - \(\dfrac{4}{5}\) . \(\dfrac{82}{7}\)= \(\dfrac{4}{5}\) .\(\dfrac{89}{7}\) - \(\dfrac{4}{5}\) . \(\dfrac{82}{7}\)= \(\dfrac{4}{5}\)\(\left(\dfrac{89}{7}-\dfrac{82}{7}\right)\)= \(\dfrac{4}{5}\). 1= \(\dfrac{4}{5}\)