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tính theo cách hợp lý
a) \(\dfrac{4}{20}+\dfrac{16}{42}+\dfrac{6}{15}+\dfrac{-3}{5}+\dfrac{2}{21}+\dfrac{-40}{21}+\dfrac{3}{20}\)
\(=\dfrac{1}{5}+\dfrac{8}{21}+\dfrac{2}{5}+\dfrac{-3}{5}+\dfrac{2}{21}+\dfrac{-40}{21}+\dfrac{3}{20}\)
\(=\left(\dfrac{1}{5}+\dfrac{-3}{5}+\dfrac{2}{5}\right)+\left(\dfrac{8}{21}+\dfrac{2}{21}+\dfrac{-10}{21}\right)+\dfrac{3}{20}\)
\(=0+0+\dfrac{3}{20}\)
\(=\dfrac{3}{20}\)
b) \(\dfrac{42}{46}+\dfrac{250}{186}+\dfrac{-2121}{2323}+\dfrac{-125125}{143143}\)
\(=\dfrac{21}{23}+\dfrac{125}{93}+\dfrac{-21}{23}+\dfrac{-125}{143}\)
\(=\left(\dfrac{21}{23}-\dfrac{-21}{23}\right)+\dfrac{125}{93}+\dfrac{-125}{143}\)
\(=0+\dfrac{125}{93}+\dfrac{-125}{143}\)
\(=\dfrac{125}{93}+\dfrac{-125}{143}\)
\(=\dfrac{17875}{13299}+\dfrac{-11625}{13299}\)
\(=\dfrac{6250}{13299}\)
\(\dfrac{42}{46}+\dfrac{250}{186}+\dfrac{-2121}{2323}+\dfrac{-125125}{143143}\)
\(=\dfrac{21}{23}+\dfrac{125}{143}+\left(\dfrac{-21}{23}\right)+\left(\dfrac{-125}{143}\right)\)
\(=\dfrac{21}{23}+\dfrac{125}{143}-\dfrac{21}{23}-\dfrac{125}{143}\)
\(=0\)
\(\dfrac{42}{46}+\dfrac{250}{186}+\dfrac{-2121}{2323}+\dfrac{-125125}{143143}\)
\(=\dfrac{21}{23}+\dfrac{125}{93}+\dfrac{-21}{23}+\dfrac{-125}{143}\)
\(=125\left(\dfrac{1}{93}-\dfrac{1}{143}\right)\)
\(=\dfrac{6250}{13299}\)
a, \(\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{-10}{21}+\frac{3}{20}\)
\(=\left(\frac{4}{20}+\frac{-3}{5}+\frac{6}{15}+\frac{3}{20}\right)\)\(+\left(\frac{16}{42}+\frac{2}{21}+\frac{-10}{21}\right)\)
\(=\left(\frac{4}{20}+\frac{-12}{20}+\frac{8}{20}+\frac{3}{20}\right)\)\(+\left(\frac{8}{21}+\frac{2}{21}+\frac{-10}{21}\right)\)
\(=\frac{3}{20}\)
b,\(\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\)
\(=\frac{21}{23}+\frac{125}{93}+\frac{-21}{23}+\frac{-125}{143}\)
\(=\left(\frac{21}{23}+\frac{-21}{23}\right)+\left(\frac{125}{93}+\frac{-125}{143}\right)\)
\(=\frac{6250}{13299}\)
`Answer:`
\(\frac{4}{20}+\frac{16}{42}+\frac{6}{15}-\left(-\frac{3}{21}\right)+\left(-\frac{10}{21}\right)+\frac{3}{20}\)
\(=\left(\frac{4}{20}+\frac{3}{20}+\frac{6}{15}\right)+\left(\frac{8}{21}+\frac{3}{21}-\frac{10}{21}\right)\)
\(=\frac{3}{4}+\frac{1}{21}\)
\(=\frac{67}{84}\)
\(\frac{42}{46}+\frac{250}{186}+\left(-\frac{2121}{2323}\right)+\left(-\frac{125125}{143143}\right)\)
\(=\frac{21}{23}+\frac{125}{93}+\left(-\frac{21}{23}\right)+\left(-\frac{125}{43}\right)\)
\(=\left(\frac{21}{23}+-\frac{21}{23}\right)+\left(\frac{125}{93}+-\frac{125}{43}\right)\)
\(=\frac{6250}{13299}\)
a: \(C=\dfrac{-1}{2}:\dfrac{-2}{3}:\dfrac{-3}{4}\)
\(=\dfrac{-1}{2}\cdot\dfrac{3}{-2}\cdot\dfrac{4}{-3}\)
\(=-\dfrac{4}{4}=-1\)
b: \(D=\dfrac{-6}{5}:\dfrac{-5}{4}:\dfrac{-4}{3}\)
\(=-\dfrac{6}{5}\cdot\dfrac{4}{5}\cdot\dfrac{3}{4}\)
\(=\dfrac{-18}{25}\)
c: \(\dfrac{4}{20}+\dfrac{16}{42}+\dfrac{6}{15}+\dfrac{-3}{5}+\dfrac{2}{21}+\dfrac{-10}{21}+\dfrac{3}{20}\)
\(=\left(\dfrac{4}{20}+\dfrac{-3}{5}+\dfrac{6}{15}\right)+\left(\dfrac{16}{42}+\dfrac{2}{21}-\dfrac{10}{21}\right)+\dfrac{3}{20}\)
\(=\left(\dfrac{1}{5}-\dfrac{3}{5}+\dfrac{2}{5}\right)+\left(\dfrac{8}{21}+\dfrac{2}{21}-\dfrac{10}{21}\right)+\dfrac{3}{20}\)
\(=\dfrac{3}{20}\)
d: \(\dfrac{42}{46}+\dfrac{250}{186}+\dfrac{-2121}{2323}+\dfrac{-125125}{143143}\)
\(=\dfrac{21}{23}+\dfrac{125}{93}+\dfrac{-21}{21}+\dfrac{-125}{143}\)
\(=\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)
a) \(\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)
\(=\frac{1}{5}+\frac{8}{21}+\frac{2}{5}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)
= \(\left(\frac{1}{5}+\frac{2}{5}+\frac{-3}{5}\right)+\left(\frac{8}{21}+\frac{2}{21}+\frac{-10}{21}\right)+\frac{3}{20}\)
\(=\frac{0}{5}+\frac{0}{21}+\frac{3}{20}\)
\(=0+0+\frac{3}{20}\)
\(=\frac{3}{20}\)
b) \(\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\)
\(=\frac{21}{23}+\frac{125}{93}+\frac{-21}{23}+\frac{-125}{143}\)
\(=\left(\frac{21}{23}+\frac{-21}{23}\right)+\frac{125}{93}+\frac{-125}{143}\)
\(=0+\frac{125}{93}+\frac{-125}{143}\)
\(=\frac{17875}{13299}+\frac{-11625}{13299}\)
\(=\frac{6250}{13299}\)
(Hình như câu b hơi sai)
\(a)\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{10}{21}+\frac{3}{20}\)
\(=\frac{1}{5}+\frac{8}{21}+\frac{2}{5}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)
\(=\left(\frac{1}{5}+\frac{2}{5}+\frac{-3}{5}\right)+\left(\frac{8}{21}+\frac{2}{21}+\frac{-10}{21}\right)+\frac{3}{20}\)
\(=0+0+\frac{3}{20}\)
\(=\frac{3}{20}\)
A.\(\dfrac{4}{20}\)+\(\dfrac{16}{42}\)+\(\dfrac{6}{15}\)+\(\dfrac{-3}{5}\)+\(\dfrac{2}{21}\)+\(\dfrac{-10}{21}\)+\(\dfrac{3}{20}\)
=\(\dfrac{1}{5}\)+\(\dfrac{8}{21}\)+\(\dfrac{6}{15}\)+\(\dfrac{-3}{5}\)+\(\dfrac{2}{21}\)+\(\dfrac{-10}{21}\)+\(\dfrac{3}{20}\)
=\((\dfrac{1}{5}\)+\(\dfrac{6}{15}\)+\(\dfrac{-3}{5}\)\()\)+\((\dfrac{8}{21}\)+\(\dfrac{2}{21}\)+\(\dfrac{-10}{21}\)\()\)+\(\dfrac{3}{20}\)
=\((\)\(\dfrac{1}{5}\)+\(\dfrac{2}{5}\)+\(\dfrac{-3}{5}\)\()\)+ 0 +\(\dfrac{3}{20}\)
= 0 + 0 +\(\dfrac{3}{20}\)
=\(\dfrac{3}{20}\)
B.\(\dfrac{42}{46}\)+\(\dfrac{250}{286}\)+\(\dfrac{-2121}{2323}\)+\(\dfrac{-125125}{143143}\)
=\(\dfrac{21}{23}\)+\(\dfrac{125}{143}\)+\(\dfrac{-21}{23}\)+\(\dfrac{-125}{134}\)
=\((\dfrac{21}{23}+\dfrac{-21}{23})+(\dfrac{125}{143}+\dfrac{-125}{134})\)
= 0 + 0
= 0
Chúc bạn Tran Mai học tốt nha
a/ \(=\frac{21}{23}+\frac{125}{143}-\frac{101.21}{101.23}-\frac{1001.125}{1001.143}=0\)
b/ \(=\frac{4}{20}+\frac{8}{21}+\frac{2}{5}-\frac{3}{5}+\frac{2}{21}-\frac{10}{21}+\frac{3}{20}=\frac{7}{20}-\frac{1}{5}=\frac{4}{20}\)
c/ \(\frac{C}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)
\(\frac{C}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{21-20}{20.21}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{20}-\frac{1}{21}\)
\(\frac{C}{2}=\frac{1}{2}-\frac{1}{21}=\frac{19}{42}\Rightarrow C=\frac{19}{21}\)
= \(\dfrac{5}{2}(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021})\)
= \(\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{5}{2}.\dfrac{100}{101}\)
= \(\dfrac{250}{101}\)
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Câu hỏi của Tran Mai - Toán lớp 6 | Học trực tuyến