Cho B = \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2006^2}\)
Chứng minh rằng B <\(\frac{334}{2007}\)
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?????????
a) Tìm giá trị lớn nhất của biểu thức: \(B=\left|3x-2\right|-\left|3x+7\right|+1\)
b) Cho \(A=\frac{10^{2006}+53}{9}\)Chứng minh rằng A là một số tự nhiên.
c) Cho \(S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\)Chứng minh rằng S không phải là số tự nhiên.
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
Chứng minh rằng : B<1
Ta có
\(\frac{1}{2^2}< \frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2}-\frac{1}{3}\)
\(........\)
\(\frac{1}{8^2}< \frac{1}{7}-\frac{1}{8}\)
\(\Rightarrow B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}=\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
Mà \(\frac{3}{8}< 1\)
\(\Rightarrow B< 1\)
Đặt A =\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}...+\frac{1}{7}-\frac{1}{8}\)
\(A=1-\frac{1}{8}< 1\)
\(\Leftrightarrow B< A< 1\)
1/ tìm a,b,c biết
(a\(^2\)+1)(b\(^2\)+2)c\(^2\)+8)-32abc=0
2/cho các số dương a,b,c,d biết:
\(\frac{a}{1+a}+\frac{b}{1+b}\frac{c}{1+c}\frac{d}{1+d}\)<=1
chứng minh rằng abcd<
3/cho abc=1 tính tổn
\(\frac{1}{1+a+ab}+\frac{1}{1+b+bc}+\frac{1}{1+c+ca}\)
4/ cho biết ã+by+cz=0 và a+b+c=\(\frac{1}{2006}\)
chứng minh rằng \(\frac{ax^2+by^2+cz^2}{bc\left(y-z\right)^2+ac\left(x-z\right)^2-ab\left(z-y\right)^2}\)=2006
Cho B=\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2016^2}\)
Chứng minh rằng B< \(\frac{1007}{6051}\)
Chứng minh rằng :
a) \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
b) \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}>48\)
a) Ta có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)
. . .
\(\frac{1}{100^2}< \frac{1}{99\cdot100}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2^2}\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\right)\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(1+1-\frac{1}{50}\right)\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{4}\cdot\frac{99}{50}=\frac{99}{200}< \frac{100}{200}=\frac{1}{2}\left(đpcm\right)\)
b) Ta có :
\(B=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}>48\)
\(\Rightarrow1-\frac{1}{4}+1-\frac{1}{9}+...+1-\frac{1}{2500}>48\)
\(\Rightarrow49-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< 49\)
Lại có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)
. . .
\(\frac{1}{50^2}< \frac{1}{49\cdot50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow\frac{1}{2^2}+...+\frac{1}{50^2}< \frac{49}{50}< 1\)
\(\Rightarrow-\left(\frac{1}{2^2}+...=\frac{1}{50^2}\right)>1\)
\(\Rightarrow49-\left(\frac{1}{2^2}+...+\frac{1}{50^2}\right)>49-1=48\)
hay \(\frac{3}{4}+\frac{8}{9}+...+\frac{2499}{2500}>48\left(đpcm\right)\)
Mik lười quá bạn tham khảo câu 3 tại đây nhé:
Câu hỏi của nguyen linh nhi - Toán lớp 6 - Học toán với OnlineMath
\(S=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\)
\(2S=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}-\frac{1}{38\cdot39}\)
\(2S=\frac{1}{2}-\frac{1}{38\cdot39}\)
\(S=\frac{1}{4}-\frac{1}{2\cdot38\cdot39}< \frac{1}{4}\)
Bài 1:
a) Cho \(b\in n\):\(b>1\). Chứng minh rằng: \(\frac{1}{b}-\frac{1}{b+1}< \frac{1}{b^2}-\frac{1}{b-1}-\frac{1}{b}\)(1)
b) Áp dụng công thức (1) chứng minh \(\frac{2}{5}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{8^2}+\frac{1}{9^2}< \frac{8}{9}\)
Bài 2. Chứng tỏ
\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{21.25}< \frac{1}{4}\)
bài 1: tính A:=\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}-\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{2}{3}-\frac{1}{2}\)
Bài 2: Cho B=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+.....+\frac{1}{49}-\frac{1}{50}\)
Chứng minh rằng: \(\frac{7}{12}< A< \frac{5}{6}\)