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![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\left(1^2+2^2+...............+2015^2\right):\left(2^2+4^2+........+4030^2\right)\)
\(P=\left(1^2+2^2+............+2015^2\right):\left[\left(1.2\right)^2+\left(2.2\right)^2+.............+\left(2.2015\right)^2\right]\)
\(P=\left(1^2+2^2+........+2015^2\right):\left(1^2.2^2+2^2.2^2+...............+2015^2.2^2\right)\)
\(P=\left(1^2+2^2+......+2015^2\right):2^2.\left(1^2+2^2+.........+2015^2\right)\)
\(P=\left(1^2+2^2+........+2015^2\right).\frac{1}{2^2.\left(1^2+2^2+..............+2015^2\right)}\)
\(P=\frac{1^2+2^2+...............+2015^2}{2^2.\left(1^2+2^2+............+2015^2\right)}=\frac{1}{2^2}=\frac{1}{4}\)
Chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
S= (-2)-(-2)2+(-2)3-(-2)4+...+(-2)2019-(-2)2020
S= -2+ 22 +(-2)3 +24 +....+(-2)2019+22020
S= -2 +(-2)3 +.....+(-2)2019 + 22 +24+....+22020
Đặt A= -2+ (-2)3+....+(-2)2019
(-2)2A= -22[-2+ (-2)3+....+(-2)2019 ]
(-2)2A= (-2)2.(-2)+ (-2)3.(-2)2+......+(-2)2. (-2)2019
4A-A= [(-2)3 + (-2)5+.....+ (-2)2021 ] - [-2+ (-2)3+....+(-2)2019 ]
3A= (-2)2021 -(-2)
3A= (-2)2021 +2
A= [(-2)2021 +2 ]:3
Đặt B= 22 +24+....+22020
22B =22 ( 22 +24+....+22020)
22B= 22.22+ 24.22+...+22.22020
4B = 24 + 26+...+22022
4B-B= (24 + 26+...+22022)-( 22 +24+....+22020)
3B= 22022-22
B= ( 22022-22):3
=> S= ( 22022-22):3 + [(-2)2021 +2 ]:3
=> S= [22022-22+(-2)2021 +2] :3
Vậy....
Ko chắc nhaa :<
Bạn https://olm.vn/thanhvien/chi5asv làm gần đúng rồi.
Sửa lại dòng 2 và 3 từ trên xuống dưới:
S = -2 - 22 + (-2)3 - 24 +...+ (-2)2019 - 22020
S = -2 + (-2)3 +...+ (-2)2019 - (22 + 24 +...+ 22020)
Sửa lại dòng 4 và dòng 5 từ dưới lên trên:
=> S = [(-2)2021 + 2] ÷ 3 - (22022 - 22) ÷ 3
=> S = [(-2)2021 + 2 - 22022 + 22] ÷ 3
=> S = 22021 + 2
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Dễ thấy \(\frac{1}{2^2}< \frac{1}{1\cdot2};...;\frac{1}{100^2}< \frac{1}{99\cdot100}\)
Do đó : \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< 1\left(đpcm\right)\)
Ta có : \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}221+321+421+...+10021<1.21+2.31+3.41+...+99.1001
=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}< 1=1−21+21−31+31−41+...+991−1001=1−1001<1
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(-17.53-21.\left(-17\right)-17.17^0\)
\(=\)\(-17.\left(53-21\right)-17.1\)
\(=\) \(-17.32-17\)
\(=\)\(-544-17\)
\(=-561\)
b, \(\left|-2^2.2^3-3^5\right|+3^5+2009^0-\left(-1\right)^{101}\)
\(=\left|-4.8-243\right|+243+1+1^{101}\)
\(=\left|-32-243\right|+243+1+1\)
\(=\left|-275\right|+243+1+1\)
\(=275+243+1+1\)
\(=518+1+1\)
\(=519+1\)
\(=520\)
:)
![](https://rs.olm.vn/images/avt/0.png?1311)
(-7)2+(-49).[ -15+(-7)4:73 ]+(-1)2014
=49+(-49).[-15+ 7]+1
=49+(-49).(-8)+1
=49+392+1
=441+1
=442
![](https://rs.olm.vn/images/avt/0.png?1311)
Dễ quá, thực hiện qui tắc bỏ dấu ngoặc được:
\(2009+2009^2+....+2009^{2009}-1-2009-...-2009^{2008}\)
\(=-1+\left(2009-2009\right)+\left(2009^2-2009^2\right)+...+\left(2009^{2008}-2009^{2008}\right)+2009^{2008}\)
\(=2009^{2008}-1\)
\(=\left(2009-1\right)\left(2009^{2007}+2009^{2008}+...+2009+1\right)\)
\(=2008\left(2009^{2007}+2009^{2008}+...+2009+1\right)\) chia hết cho 2008
=> ĐPCM
Chứng Minh Rằng: (2009+20092+20093+20094+...+20092009)-(1+2009+20092+20093+...+20092008) chia hết cho 2008.
Đặt A=2009+20092+20093+20094+...+20092009, B=1+2009+20092+20093+20094+...+20092008
Ta có:
+)A=2009+20092+20093+20094+...+20092009
2009A= 20092+20093+20094+...+20092010
2009A-A=(20092+20093+20094+...+20092010)-(2009+20092+20093+20094+...+20092009)
2008A=20092010- 2009
=> A=(20092010- 2009)/2008
=> A chia hết cho 2008.
B=1+2009+20092+20093+20094+...+20092008
2009B=2009+20092+20093+20094+...+20092010
2009B-B=(2009+20092+20093+20094+...+20092010)-(1+2009+20092+20093+20094+...+20092009)
2008B=20092010-1
=>B=(20092010-1)/2008
=>B chia hết cho 2008
=> A-B chia hết cho 2008.
=> ĐPCM
Đặt tử số là A
Ta có: 2A = 2 + 22 + 23 + 24+...+22009
=> A = 2A - A= 22009 - 1
=> B = \(\frac{2^{2009}-1}{1-2^{2009}}\) = -1
mik cảm ơn bn :D