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KT
4 tháng 2 2018
\(\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{9}\right)\times\left(1-\frac{1}{16}\right)\times\left(1-\frac{1}{25}\right)\times\left(1-\frac{1}{36}\right)\)
\(=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times\frac{24}{25}\times\frac{36}{36}\)
\(=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times\frac{3.5}{4.4}\times\frac{4.6}{5.5}\times\frac{5.7}{6.6}\)
\(=\frac{1.2.3.4.5}{2.3.4.5.6}\times\frac{3.4.5.6.7}{2.3.4.5.6}\)
\(=\frac{1}{6}\times\frac{7}{2}\)
\(=\frac{7}{12}\)
4 tháng 2 2018
(1-1/4)×(1-1/9)×(1-1/16)×(1-1/25)×(1-1/36)
=(4/4-1/4)×(9/9-1/9)×(16/16-1/16)×(25/25-1/25)×(36/36-1/36)
=3/4×8/9×15/16×24/25×35/36
=1×3×2×4×3×5×4×6×5×7/2×2×3×3×4×4×5×5×6×6
=(1×2×3×4×5)×(3×4×5×6×7)/(2×3×4×5×6)×(2×3×4×5×6)
=1/6×7/2
=7/12
CH
0
VN
5
25 tháng 8 2018
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).\left(1-\frac{1}{5}\right)\)
\(=\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}\)
\(=\frac{2}{6}\)
\(=\frac{1}{3}\)
25 tháng 8 2018
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).\left(1-\frac{1}{5}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)
\(=\frac{1}{5}\)
\(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).....\left(1-\frac{1}{36}\right)\)
\(=\frac{3}{4}.\frac{8}{9}....\frac{35}{36}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}.\frac{5.7}{6.6}\)
\(=\frac{7}{2.6}=\frac{7}{12}\)
P/S: dấu "." là dấu nhân nhé
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 7
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 7
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (\(\dfrac{2}{2}\) + \(\dfrac{3}{3}\) + \(\dfrac{4}{4}\) + \(\dfrac{5}{5}\)+ \(\dfrac{6}{6}+\dfrac{7}{7}+\dfrac{8}{8}\) + \(\dfrac{10}{10}\))
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (1 + 1 +1 + 1+ 1+ 1+ 1 +1)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x 1 x 8
= \(\dfrac{1}{2}\) + \(\)\(\dfrac{1}{2}\) x 8
= \(\dfrac{1}{2}\) + 4
= \(\dfrac{9}{2}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 9
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + 8
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 9
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{9}{18}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)
= \(\dfrac{1}{2}\) x 10
= 5
BB
2
6 tháng 4 2021
Ta có: \(\dfrac{1}{5}+\dfrac{4}{10}+\dfrac{9}{15}+\dfrac{16}{20}+\dfrac{25}{25}+\dfrac{36}{30}+\dfrac{49}{35}+\dfrac{64}{40}+\dfrac{81}{45}\)
\(=\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5}+\dfrac{5}{5}+\dfrac{6}{5}+\dfrac{7}{5}+\dfrac{8}{5}+\dfrac{9}{5}\)
\(=\dfrac{45}{5}=9\)
11 tháng 4 2017
\(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right).\left(1-\frac{1}{25}\right).\left(1-\frac{1}{36}\right)\)
\(=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.\frac{35}{36}=\frac{3.8.15.24.35}{4.9.16.25.36}=\frac{1.3.2.4.3.5.4.6.5.7}{2.2.3.3.4.4.5.5.6.6}\)
\(=\frac{\left(1.2.3.4.5\right).\left(3.4.5.6.7\right)}{\left(2.3.4.5.6\right).\left(2.3.4.5.6\right)}=\frac{1.7}{2.2}=\frac{7}{4}\)
DT
3
6 tháng 11 2016
1+4+9+16+25+....+100
Theo quy luật:
12+22+32+42+52+...+102
=385