tính nhanh
1+1+1+1+1+2+2+2+3+3+3
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Những câu hỏi liên quan
15 tháng 2 2022
\(\dfrac{1}{7}\times\dfrac{21}{8}-\dfrac{3}{8}\times\dfrac{1}{7}-\dfrac{1}{7}\times\dfrac{2}{8}\\ =\dfrac{1}{7}\times\left(\dfrac{21}{8}-\dfrac{3}{8}-\dfrac{2}{8}\right)\\ =\dfrac{1}{7}\times\dfrac{16}{8}\\ =\dfrac{1}{7}\times2\\ =\dfrac{2}{7}\)
TD
15 tháng 2 2022
\(\dfrac{1}{7}\times\dfrac{21}{8}-\dfrac{3}{8}\times\dfrac{1}{7}-\dfrac{1}{7}\times\dfrac{2}{8}\)
=\(\dfrac{1}{7}\times\left(\dfrac{21}{8}-\dfrac{3}{8}-\dfrac{2}{8}\right)\)
=\(\dfrac{1}{7}\times2\)
=\(\dfrac{2}{7}\)
30 tháng 3 2022
\(\dfrac{1}{3}\left(\dfrac{2}{7}+\dfrac{4}{7}+\dfrac{1}{7}\right)=\dfrac{1}{3}\times1=\dfrac{1}{3}\)
8 tháng 6 2023
2:
=1-1+1-1=0
3:
a: =>34*(100+1)/2:a=17
=>a=101
b: =>5/3(x-1/2)=5/4
=>x-1/2=5/4:5/3=3/4
=>x=5/4
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
8 tháng 6 2023
1a, \(\dfrac{2005}{2001}\) = 1+\(\dfrac{4}{2001}\); \(\dfrac{2009}{2005}\)=1+\(\dfrac{4}{2005}\)vì\(\dfrac{4}{2001}\)>\(\dfrac{4}{2005}\)nên\(\dfrac{2005}{2001}\)>\(\dfrac{2009}{2005}\)
1b,\(\dfrac{1313}{1515}\)=\(\dfrac{1313:101}{1515:101}\)= \(\dfrac{13}{15}\); \(\dfrac{131313}{151515}\)=\(\dfrac{131313:10101}{151515:10101}\)=\(\dfrac{13}{15}\)
Vậy \(\dfrac{13}{15}\)=\(\dfrac{1313}{1515}\)=\(\dfrac{131313}{151515}\)
NK
17 tháng 4 2023
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\)
\(=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(=\dfrac{1}{1}\cdot\dfrac{1}{2}+\dfrac{1}{2}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{1}{5}+\dfrac{1}{5}\cdot\dfrac{1}{6}+\dfrac{1}{6}\cdot\dfrac{1}{7}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(=\dfrac{1}{1}-\dfrac{1}{7}=\dfrac{7}{7}-\dfrac{1}{7}=\dfrac{6}{7}\)
NT
3
9 tháng 4 2021
Ta có: \(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{9702}\)
\(=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{98\cdot99}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}\)
\(=\dfrac{32}{99}\)
9 tháng 4 2021
Giải:
1/12+1/20+1/30+...+1/9702
=1/3.4+1/4.5+1/5.6+...+1/98.99
=1/3-1/4+1/4-1/5+1/5-1/6+...+1/98-1/99
=1/3-1/99
=32/99
Chúc bạn học tốt!
1+1+1+1+1+2+2+2+3+3+3
=1*5+2*3+3*3
=5+6+9
=20
20
k mk nha ^^