1+2+21+22+....+22019
Giúp mình nhé mình cần gấp
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Ta có: \(\frac{1}{21}>\frac{1}{30};\frac{1}{22}>\frac{1}{30};\ldots;\frac{1}{30}=\frac{1}{30}\)
Do đó: \(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+\cdots+\frac{1}{30}=\frac{10}{30}=\frac13\) (1)
Ta có: \(\frac{1}{31}>\frac{1}{40};\frac{1}{32}>\frac{1}{40};\ldots;\frac{1}{40}=\frac{1}{40}\)
Do đó: \(\frac{1}{31}+\frac{1}{32}+\cdots+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+\cdots+\frac{1}{40}=\frac{10}{40}=\frac14\) (2)
Ta có: \(\frac{1}{41}>\frac{1}{50};\frac{1}{42}>\frac{1}{50};\ldots;\frac{1}{50}=\frac{1}{50}\)
Do đó: \(\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{50}>\frac{1}{50}+\frac{1}{50}+\cdots+\frac{1}{50}=\frac{10}{50}=\frac15\) (3)
Từ (1),(2),(3) suy ra \(\left(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+\cdots+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{50}\right)>\frac13+\frac14+\frac15=\frac{47}{60}\)
Ta có: \(\frac{1}{21}>\frac{1}{30};\frac{1}{22}>\frac{1}{30};\ldots;\frac{1}{30}=\frac{1}{30}\)
Do đó: \(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+\cdots+\frac{1}{30}=\frac{10}{30}=\frac13\) (1)
Ta có: \(\frac{1}{31}>\frac{1}{40};\frac{1}{32}>\frac{1}{40};\ldots;\frac{1}{40}=\frac{1}{40}\)
Do đó: \(\frac{1}{31}+\frac{1}{32}+\cdots+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+\cdots+\frac{1}{40}=\frac{10}{40}=\frac14\) (2)
Ta có: \(\frac{1}{41}>\frac{1}{50};\frac{1}{42}>\frac{1}{50};\ldots;\frac{1}{50}=\frac{1}{50}\)
Do đó: \(\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{50}>\frac{1}{50}+\frac{1}{50}+\cdots+\frac{1}{50}=\frac{10}{50}=\frac15\) (3)
Từ (1),(2),(3) suy ra \(\left(\frac{1}{21}+\frac{1}{22}+\cdots+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+\cdots+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+\cdots+\frac{1}{50}\right)>\frac13+\frac14+\frac15=\frac{47}{60}\)
a) C1 : \(\left(\frac{1}{2}+\frac{1}{3}\right)\times\frac{2}{5}=\left(\frac{3}{6}+\frac{2}{6}\right)\times\frac{2}{5}\)
\(=\frac{5}{6}\times\frac{2}{5}\)
\(=\frac{10}{30}=\frac{1}{3}\)
C2 : \(\left(\frac{1}{2}+\frac{1}{3}\right)\times\frac{2}{5}=\frac{1}{2}\times\frac{2}{5}+\frac{1}{3}\times\frac{2}{5}\)
\(=\frac{1}{5}+\frac{2}{15}\)
\(=\frac{3}{15}+\frac{2}{15}\)
\(=\frac{5}{15}=\frac{1}{3}\)
b) C1 : \(\frac{3}{5}\times\frac{17}{21}+\frac{17}{21}\times\frac{2}{5}=\frac{17}{35}+\frac{34}{105}\)
\(=\frac{51}{105}+\frac{34}{105}\)
\(=\frac{85}{105}=\frac{17}{21}\)
C2 : \(\frac{3}{5}\times\frac{17}{21}+\frac{17}{21}\times\frac{2}{5}=\frac{17}{21}\times\left(\frac{3}{5}+\frac{2}{5}\right)\)
\(=\frac{17}{21}\times1=\frac{17}{21}\)
`45-21/18xx2/15`
`=45-7/6xx2/15`
`=45-7/45`
`=2025/45-7/45`
`=2018/45`
giả sử số cần tìm là A , ta có
A=1+2+21+22+....+22019
2A=2+ 21+22+23+....+22020
2A-A= (2+21+22+23+....+22020) - (1+2+21+22+....+22019)
A=22020 - 2
Đăt A= 1+2+22+......+22019
2A=2(1+2+22+.....+22019)
2A=2+22+23+....+22020
2A-A=(2+22+23+.....+22020)-(1+2+22+....+22019)
A =2+22+23+....+22020-1-2-22-....-22019
A=22020-1