So sánh A=1+21+22+23+24+25+...+250+251 với B=225
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Số số hạng của tổng A là : \(\dfrac{30-21}{1}+1=10\left(sh\right)\)
`=>A=\underbrace{1/21+1/22+...+1/30}_{10sh}>\underbrace{1/30+1/30+1/30+...+1/30}_{10sh}`
`=>A>(1)/(30).10`
`=>A>10/30`
`=>A>1/3`
`=>đpcm`
![](https://rs.olm.vn/images/avt/0.png?1311)
1−2−3+4+5−6−7+8+...+21−22−23+24+25
= (1 - 2 - 3 + 4) + (5 - 6 - 7 + 8) + ... + (21 - 22 - 23 + 24) + 25=(1−2−3+4)+(5−6−7+8)+...+(21−22−23+24)+25
= 0 + 0 + ... + 0 + 25=0+0+...+0+25
= 25
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Vì: \(\frac{3}{21}=\frac{3}{21}\)
\(\frac{3}{22}\) < \(\frac{3}{21}\)
\(\frac{3}{23}\) < \(\frac{3}{21}\)
\(\frac{3}{24}\)<\(\frac{3}{21}\)
\(\frac{3}{25}\)< \(\frac{3}{21}\)
.....
\(\frac{2}{29}\)<\(\frac{3}{21}\)
\(\frac{2}{30}\)<\(\frac{3}{21}\)
Nên \(\frac{3}{21}+\frac{3}{22}+\frac{3}{23}+\frac{3}{24}+\frac{3}{25}+...+\frac{3}{29}+\frac{3}{30}\) < \(\frac{3}{21}.10\)
Ta có: \(\frac{3}{21}.10\) = \(\frac{10}{7}\)
Mà \(\frac{10}{7}\) < \(\frac{3}{2}\)
=>\(\frac{3}{21}+\frac{3}{22}+\frac{3}{23}+\frac{3}{24}+\frac{3}{25}+...+\frac{3}{29}+\frac{3}{30}\) < \(\frac{3}{2}\)
Vậy E < M
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: 2^25 - 2^24 + 2^23 = 2^23 . (2^2-2+1) = 2^23.3
2^23-2^22 + 2^21 =2^21.(2^2-2+1) = 2^21.3
=> 2^23.3 > 2^21.3
=> 2^25 - 2^24 + 2^23 > 2^23 - 2^22 + 2^21
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a) ( 72 - 8x9 ) : ( 20 + 21 + 22 + 23 + 24 + 25 )
= ( 72 - 72 ) : ( 20 + 21 + 22 + 23 + 24 + 25 )
= 0 : ( 20 + 21 + 22 + 23 + 24 + 25 )
= 0
b) ( 500 x 9 - 250 x 18 ) x ( 1 + 2 + 3 + ... + 9 )
= ( 250 x 2 x 9 - 250 x 18 ) x ( 1 + 2 + 3 + ... + 9 )
= ( 250 x 18 - 250 x 18 ) x ( 1 + 2 + 3 + ... + 9 )
= 0 x ( 1 + 2 + 3 + ... + 9 )
= 0
c ) ( 11 + 13 + 15 + ... + 19 ) x ( 6 x 8 - 48 )
= ( 11 + 13 + 15 + ... + 19 ) x ( 48 - 48 )
= ( 11 + 13 + 15 + ... + 19 ) x 0
= 0
a) ( 72 - 8x9 ) : ( 20 + 21 + 22 + 23 + 24 + 25 )
= ( 72 - 72 ) : ( 20 + 21 + 22 + 23 + 24 + 25 )
= 0 : ( 20 + 21 + 22 + 23 + 24 + 25 )
= 0
![](https://rs.olm.vn/images/avt/0.png?1311)
A>B vì 251>225 mà các số trong A đều lớn hơn 0
thank