98 nhân 98 + 98 + 98
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NT
0
VN
0
2C
2
15 tháng 4 2019
\(\frac{1}{2\cdot4}+\frac{1}{6\cdot8}+...+\frac{1}{96\cdot98}+\frac{1}{98\cdot100}\)
\(=\frac{1}{2}\left[\frac{2}{2\cdot4}+\frac{2}{6\cdot8}+...+\frac{2}{96\cdot98}+\frac{2}{98\cdot100}\right]\)
\(=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right]\)
\(=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{100}\right]=\frac{1}{2}\left[\frac{50}{100}-\frac{1}{100}\right]=\frac{1}{2}\cdot\frac{49}{100}=\frac{49}{200}\)
S
15 tháng 4 2019
\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}...+\frac{1}{96.98}+\frac{1}{98.100}\)
\(=\frac{1}{2}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{96.98}+\frac{2}{98.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{49}{100}\)
\(=\frac{49}{200}\)
~Học tốt~
LL
1
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
\(C-D=\dfrac{\left(98^{99}+1\right)\left(98^{88}+1\right)-\left(98^{89}+1\right)\left(98^{98}+1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)
\(=\dfrac{98^{187}+98^{99}+98^{88}+1-98^{197}-98^{89}-98^{98}-1}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)
\(=\dfrac{98^{99}-98^{98}+98^{88}-98^{89}}{\left(98^{89}+1\right)\left(98^{88}+1\right)}=\dfrac{98^{98}\left(98-1\right)-98^{88}\left(98-1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}\)
\(=\dfrac{97.98^{98}-97.98^{88}}{\left(98^{89}+1\right)\left(98^{88}+1\right)}=\dfrac{97.98^{88}\left(98^{10}-1\right)}{\left(98^{89}+1\right)\left(98^{88}+1\right)}>0\)
\(\Rightarrow C>D\)
A
0
T
6
13 tháng 12 2020
\(\left(9^{98}\cdot80+9^{98}\right):9^{100}\)
\(=\frac{9^{98}\cdot80+9^{98}}{9^{100}}\)
\(=\frac{9^{98}\cdot\left(80+1\right)}{9^{100}}\)
\(=\frac{9^{98}\cdot81}{100}\)
\(=\frac{9^{98}\cdot9^2}{9^{100}}\)
\(=\frac{9^{100}}{9^{100}}=1\)
NH
0
28812
98 x 98 + 98 + 98 = 9800