chứng tỏ rằng \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2010.2013}< \frac{1}{3}\)
Những câu hỏi liên quan
Cho S=\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.....+\frac{3}{40.43}+\frac{3}{43.46}.\)
Hãy chứng tỏ rằng S<1
\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
Có \(1-\frac{1}{46}< 1\)
\(\Rightarrow S< 1\)
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nhan xet:3/1.4=1/1-1/4
3/4.7=1/4-1/7
3/7.10=1/7-1/10
.....................
3/40.43=1/40-1/43
3/43.46=1/43-1/46
S=1/1-1/3+1/3-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46
S=1/1-1/46
S=46/46-1/46
S=45/46<1
vay s<1
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S=\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
chứng tỏ rằng S<1
S= 1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+|1/43-1/46
S= 1-1/46
S= 45/46<1
vậy S<1
duyệt đi
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S= 1- 1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46
S= 1+ (1/4-1/4)+(1/7-1/7)+...+(1/43-1/43)-1/46
S= 1-1/46= 45/46<1
Suy ra S<1
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Bài 5: Cho S = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\). Hãy chứng tỏ rằng S < 1.
Cho \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\).Hãy chứng tỏ rằng S < 1
Có \(S=\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{3}{43.46}\)
Sẽ có: \(S=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+....+\frac{1}{43.46}\)
\(S=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+........+\frac{1}{43}-\frac{1}{46}\)
\(S=\frac{1}{1}-\left(-\frac{1}{4}+\frac{1}{4}\right)+\left(-\frac{1}{7}+\frac{1}{7}\right)+\left(-\frac{1}{10}+\frac{1}{10}\right)+....+\left(-\frac{1}{43}+\frac{1}{43}\right)-\frac{1}{46}\)
\(S=\frac{1}{1}-0+0+0+0+......+0+0-\frac{1}{46}\)
\(S=\frac{1}{1}-\frac{1}{46}=\frac{45}{46}\)
Vì có: \(\frac{45}{46}<1\) nên \(S<1\)
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\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}\)
\(S=\frac{45}{46}\)
=> \(\frac{45}{46}<\frac{46}{46}=1\)
=> S<1
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Cho \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\) . Hãy chứng tỏ rằng \(S< 1\)
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
ta có
\(\frac{3}{1.4}=1-\frac{1}{4}\)
\(\frac{3}{4.7}=\frac{1}{4}-\frac{1}{7}\)
.....
\(\frac{3}{43.46}=\frac{1}{43}-\frac{1}{46}\)
( mình nghĩ cậu chưa đc làm dạng như này nên ghi ra , lần sau có gặp mà biết cách làm r thì bỏ bước trên đi cx đc nha)
\(=>S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(=>S=1-\frac{1}{46}< 1\)(dpcm
bài 1,Cho \(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\).Hãy chứng tỏ rằng B>1.
bài 2,Cho \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\).Hãy chứng tỏ rằng S<1.
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}< 1\)
Vậy \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}< 1\)
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Cho S=\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.....+\frac{3}{n.\left(n+3\right)}\) với n thuộc N*
Chứng tỏ rằng S<1
=>S= 1- 1/4 + 1/4 -1/7 + 1/7 - 1/10 +...+ 1/n - 1/(n+3)
=>S= 1- 1/(n+3)
=>S + 1/(n+3) = 1
=>S<1
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\(A=\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-....-\frac{1}{2011.2014}\)
\(A=\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-...-\frac{1}{2011.2014}\)
\(A=\frac{1}{1.4}-\left(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\right)\)
Đặt \(B=\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2011.2014}\)
\(B=\frac{1}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{2011.2014}\right)\)
\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2011}-\frac{1}{2014}\right)\)
\(B=\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{2014}\right)\)
\(B=\frac{1}{3}.\frac{1005}{4028}=\frac{335}{4028}\)
\(A=\frac{1}{4}-\frac{335}{4028}=\frac{168}{1007}\)
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A = \(\frac{1}{1.4}-\frac{1}{4.7}-\frac{1}{7.10}-...-\frac{1}{2011.2014}\)
A = 1 + \(\frac{1}{4}\) - \(\frac{1}{4}\) + \(\frac{1}{7}\) - \(\frac{1}{7}\) + \(\frac{1}{10}\) -....- \(\frac{1}{2011}\) + \(\frac{1}{2014}\)
A = 1 + \(\frac{1}{2014}\) = \(\frac{2015}{2014}\)
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(\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+..........+\frac{1}{97.100}=\frac{0.33.x}{2009}\))
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\Rightarrow\frac{99}{100}=\frac{0.33.x}{2009}\)
\(\Rightarrow100.0.33.x=99.2009\)
\(\Rightarrow0x=198891\Rightarrow\)không có GT x thỏa mãn
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