so sánh: 1/4^2+1/6^2+1/8^2+...+1/2006^2 với 334/2007
Những câu hỏi liên quan
cho B = 1/4^2 +1/6^2+1/8^2 +...+1/2006^2 chứng minh B <334/2007
Help me!!!!!!!!!!!!!!
cho B = 1/4^2 + 1/6^2 +1/ 8^2 + ... + 1/2006^2. Chứng minh B< 334/2007
D= 1/42+1/62+1/82+...+1/20062. CMR: D < 334/2007
Cho B=\(\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{2006^2}\). Chứng minh : B<\(\dfrac{334}{2007}\)
=>B=\(\dfrac{1}{4.4}+\dfrac{1}{6.6}+\dfrac{1}{8.8}+...+\dfrac{1}{2006.2006}\)
=>B<\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{2005.2007}\)
=>B<\(\dfrac{2}{2}.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{2005.2007}\right)\)
=>B<\(\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2005.2007}\right)\)
=>B<\(\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2005}-\dfrac{1}{2007}\right)\)
=>B<\(\dfrac{1}{2}.\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{5}+...+\dfrac{1}{2005}-\dfrac{1}{2005}-\dfrac{1}{200}\right)\)(xin lỗi, đoạn cuối (chỗ 200 í )là 2007 nhá
=>B<\(\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{2007}\right)\)
=>B<\(\dfrac{1}{2}.\dfrac{668}{2007}\)
=>B<\(\dfrac{1.668}{2.2007}\)
=>B<\(\dfrac{1.668:2}{2.2007:2}\)
=>B<\(\dfrac{334}{2007}\)
Tick cho tôi nha :D
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Cho B =\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+......+\frac{1}{2006^2}\) Chứng minh : B < \(\frac{334}{2007}\)
Ta thấy : \(\frac{1}{4^2}< \frac{1}{4.5};\frac{1}{6^2}< \frac{1}{5.6};...;\frac{1}{2006^2}< \frac{1}{2005.2006}\)
\(\Rightarrow B=\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{2006^2}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{2005.2006}\)
\(\Leftrightarrow B< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(\Leftrightarrow B< \frac{1}{4}-\frac{1}{2006}=\frac{1001}{4012}\)
Mà \(\frac{1001}{4012}< \frac{334}{2007}\Rightarrow B< \frac{334}{2007}\)
\(B< \frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2006.2008}\)
\(2B< \frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2006.2008}=\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2006}-\frac{1}{2008}=\frac{1}{4}-\frac{1}{2008}=\frac{501}{2008}\)\(B< \frac{501}{4016}< \frac{501}{4014}< \frac{668}{4014}=\frac{334}{2007}\)
Vậy:.....
1/6^2 < 1/5.6?????????
Cho B= 1/4^2 +1/6^2+ .....+1/2006^2
Chứng tỏ B< 334/2007
Cho \(B=\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+..................+\dfrac{1}{2006^2}\). Chứng minh rằng \(B< \dfrac{334}{2007}\)
Help me!!!!!!!!!!!!!!!
Bài1:So sánh
A=2006/987654321+2007/246813579
B=2007/987654321+2006/246813579
b)1965/1976 và 1973/1975
Bài2:Tìm x
a)3 - (5 và 3/8 + x - 7 và 5/24):6 và 2/3=2
b) (1/1*2+1/2*3+1/3*4+1/4*5+1/5*6)*10 - x=0
Bài3:Tính nhanh
A=1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256
1. 2006/987654321 + 2007/246813579 = 2007/246813579 + 2006/987654321
=>
2.
3 - (5.3/8 + X - 7 . 5/24) : 6 . 2/3 =2
3 - (15/8 + X - 35/24) : 4 = 2
3 - (15/8 + X - 35/24) = 2 . 4
3 - (15/8 + X - 35/24) = 8
15/8 + X - 35/24 = 3 - 8
15/8 + X - 35/24 = -5
15/8 + X = -5 + 35/24
15/8 + X = -85/24
X = -85/24 - 15/8
X = -65/12
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30 tháng 1 2022 lúc 16:45
a,tính tổng : \(S=\dfrac{27+4500+135+550+2}{2+4+6+...+14+16+18}\)
b, So sánh : \(A=\dfrac{2006^{2006}+1}{2006^{2007}+1}v\text{à }B=\dfrac{2006^{2005}+1}{2006^{2006}+1}\)
- Mình dùng cách lớp 8 để làm câu b được không :)?
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- Tham khảo câu b:
https://olm.vn/hoi-dap/tim-kiem?q=+++++++++++A=2006%5E2005+1/2006%5E2006+1B=2006%5E2006+1/2006%5E2007+1so+s%C3%A1nh+A+v%C3%A0+B&id=520258
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