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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta thấy :
2 < 3
200 < 300
\(\Rightarrow2^{200}< 3^{300}\)
P/s : ez mà
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: 4200 = (42)100 = 16100
2400 = (24)100 = 16100
Vì 16100 = 16100 nên 4200 = 2400
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4^{50}=\left(2^2\right)^{50}=2^{100}\)
\(8^{23}=\left(2^3\right)^{23}=2^{69}\)
Vì: \(100>69\Rightarrow2^{100}>2^{69}\Rightarrow4^{50}>8^{23}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)=\frac{189}{760}\)
Đặt \(B=\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{19.20}=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{3}+...+\frac{3}{19}-\frac{3}{20}\)
\(=3-\frac{3}{20}=\frac{57}{20}\)
\(D=A-B=\frac{189}{760}-\frac{57}{20}=-\frac{1977}{760}\)
Gọi \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)là A
\(\frac{3}{1.2}-\frac{3}{2.3}-...-\frac{3}{19.20}\)là B
\(A=\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}.\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\right]\)
\(A=\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\right]\)
\(A=\left[\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\right]\)
\(A=\left[\frac{1}{2}.\left(1-\frac{1}{20}\right)\right]\)
\(A=\frac{1}{2}.\frac{19}{20}\)
\(A=\frac{19}{40}\)
\(B=\frac{3}{1.2}-\frac{3}{2.3}-...-\frac{3}{19.20}\)
\(B=\left(\frac{3}{1.2}+\frac{3}{2.3}+...+\frac{3}{19.20}\right)\)
\(B=\left[3.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{19.20}\right)\right]\)
\(B=\left[3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{2}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\right]\)
\(B=\left[3.\left(\frac{19}{20}\right)\right]\)
\(B=\frac{57}{20}\)
Vậy A - B = \(\frac{19}{40}-\frac{57}{20}\)
\(=-\frac{95}{40}=-\frac{19}{8}\)
Nếu đúng thì k nha
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
- 823 < 923 = (32)23 = 346
- 2716 = (32)16 = 332
- Vì 332 < 346 => 2716 < 823 .
Mong bạn sẽ chọn câu trả lời của mình ! Cảm ơn .
\(8^{23}< 9^{23}=\left(3^2\right)^{23}=3^{2.23}=3^{46}\)
\(27^{16}=\left(3^3\right)^{16}=3^{3.16}=3^{48}\)
Do \(46< 48\)nên \(3^{46}< 3^{48}\)
hay \(8^{23}< 27^{16}\)
hk tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
a) |2004|=2004
|-2005|=2005
|-9|=9
|8|=8
b) Ta có: |4|=4
|7|=7
Vì 4<7 nên |4|<|7|
Ta có: |-2|=2
|-5|=5
Vì 2<5 nên |-2|<|-5|
Ta có: |-3|=3
|8|=8
Vì 3<8 nên |-3|<|8|
Vậy...........
a ) Giá trị tuyệt đối của :
2004 là 2004 ; - 2005 = 2005 ; -9 = 9 ; 8 = 8 .
b ) Ta có :
/4/ = 4 ; /7/ = 7 . Vì 4 < 7 nên /4/ < /7/ .
/-2/ = 2 ; /-5/ = 5 . Vì 2 < 5 nên /-2/ < /-5/ .
/-3/ = 3 ; /8/ = 8 . Vì 3 < 8 nên /-3/ < /8/ .
![](https://rs.olm.vn/images/avt/0.png?1311)
A= (1000+1)x1000:2=500500
B=1.2.3.4........11=39916800
=>A<B
tich ủng hộ nha