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BB
2
6 tháng 4 2021
Ta có: \(\dfrac{1}{5}+\dfrac{4}{10}+\dfrac{9}{15}+\dfrac{16}{20}+\dfrac{25}{25}+\dfrac{36}{30}+\dfrac{49}{35}+\dfrac{64}{40}+\dfrac{81}{45}\)
\(=\dfrac{1}{5}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{4}{5}+\dfrac{5}{5}+\dfrac{6}{5}+\dfrac{7}{5}+\dfrac{8}{5}+\dfrac{9}{5}\)
\(=\dfrac{45}{5}=9\)
BB
2
28 tháng 3 2020
1/5+4/10+9/15+16/20+25/25+36/30+49/35+64/40+81/45 = 1/5 + 2/5 + 3/5 + 4/5 + 1 + 6/5 + 7/5 + 8/5 + 9/5 = (1/5 + 9/5) + (2/5 + 8/5) + (3/5 + 7/5) + (4/5 + 6/5) + 1 = 2 + 2 + 2 + 2 + 1 = 9
NV
1
TM
1
15 tháng 5 2019
\(\frac{1}{5}+\frac{4}{10}+\frac{9}{15}+\frac{16}{20}+1+\frac{36}{30}+\frac{49}{35}+\frac{64}{40}+\frac{81}{45}\)
\(=\left(\frac{1}{5}+\frac{81}{45}\right)+\left(\frac{4}{10}+\frac{49}{35}\right)+\left(\frac{9}{15}+\frac{49}{35}\right)+\left(\frac{16}{20}+\frac{36}{30}\right)+1\)
\(=2+2+2+2+1\)
\(=2\times4+1\)
\(=9\)
~ Hok tốt ~
16 tháng 8 2016
1/5+4/10+9/15+25/25+36/30+49/35+64/40+81/45
=1/5 + (4/10:2) + (9/15:3)+25/25 + (36/30:6)+ (49/35:7) + (64/40:8) + (81/45:9)
=1/5 + 1/5 + 1/5 + 1 + 1/5 + 1/5 + 1/5 + 1/5
=1/5 x 7 + 1
=12/5
3 tháng 8 2015
\(\frac{1}{5}+\frac{4}{10}+...+\frac{81}{45}=\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+...+\frac{9}{5}=\frac{1+2+3+...+9}{5}=\frac{45}{5}=9\)
3 tháng 8 2015
1/5 + 4/10 + 9/15 + 16/20 + 25/25 + 36/30 + 49/35 + 64/40 + 81/45
=1/5 + 2/5 + 3/5 + 4/5 + 5/5 + 6/5 + 7/5 + 8/5 + 9/5
=45/5 = 9
IA
19 tháng 2 2020
\(a,=\left(891+109\right)+359\)
\(=1000+359\)
\(=1359\)
\(b=\left(\frac{19}{11}+\frac{3}{11}\right)+\frac{5}{13}\)
\(=2+\frac{5}{13}=\frac{31}{13}\)
\(c,=\frac{1}{5}\left(1+2+3+4+5+6+7+8+9\right)\)
\(=\frac{1}{5}\cdot45=9\)
19 tháng 2 2020
a)=891+359+109
=(891+109)+35
=1000+359=1359
b)=19/11+5/13+3/11
=(19/11+3/11)+5/13
=2+5/13=31/13
c)=1/5+2/5+3/5+4/5+5/5+6/5+7/5+8/5+9/5
=(1/5+9/5)+(2/5+8/5)+(3/5+7/5)+(4/5+6/5)+5/5
=1+1+1+1+1=5
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
1/5 + 4/10 +9/15 + 16/20 + 25/25 + 36/30 + 49/35 + 64/40 + 81/45
=1/5 + 2/5 + 3/5 +...+ 9/5
=45/5=9
\(\frac{1}{5}+\frac{4}{10}+\frac{9}{15}+\frac{16}{20}+\frac{25}{25}+\frac{36}{30}+\frac{49}{35}+\frac{64}{40}+\frac{81}{45}\)
\(=\frac{1}{5}+\frac{1}{5}+\frac{3}{5}+\frac{4}{5}+\frac{5}{5}+\frac{6}{5}+\frac{7}{5}+\frac{8}{5}+\frac{9}{5}\)
\(=\frac{\left(9+1\right)\cdot9:2}{5}=\frac{45}{5}=9\)