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S
1
D
18 tháng 4 2017
Ta có: \(\frac{3}{1.4}+\frac{3}{4.7}+......+\frac{3}{40.43}+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}\)
Vì \(\frac{1}{46}>0\Rightarrow1-\frac{1}{46}< 1\)
Vậy \(\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{43.46}< 1\)
NN
2
Q
23 tháng 4 2017
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\) < 1
\(S=3\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{40.43}+\frac{1}{43.46}\right)\)
\(S=3.\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\right)\)
\(\Rightarrow S=1-\frac{1}{46}\Rightarrow S< 1\left(đpcm\right)\)
23 tháng 4 2017
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
= \(1-\frac{1}{46}< 1\)
\(\Rightarrow S< 1\left(đpcm\right)\)
TS
2
AK
1 tháng 5 2018
\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}< 1\)
Vậy \(S< 1\)
Chúc bạn học tốt !!!
HT
6
LM
3 tháng 4 2017
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{43\cdot46}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{46}=1-\frac{1}{46}< 1\)
\(\left(\frac{3}{a\cdot\left(a+3\right)}=\frac{a+3-3}{a\cdot\left(a+3\right)}=\frac{1}{a}-\frac{1}{a+3}\right)\)
3 tháng 4 2017
\(S=\frac{3}{1\times4}+\frac{3}{4\times7}+...+\frac{3}{43\times46}\)
\(3S=3-\frac{3}{4}+\frac{3}{4}-\frac{3}{7}+...+\frac{3}{43}-\frac{3}{46}\)
\(3S=3-\frac{3}{46}\)
\(3S=\frac{135}{46}\)
\(S=\frac{45}{46}< 1\)
Vậy ra có điều phải chứng minh
NY
3
10 tháng 5 2019
\(50\%.\frac{4}{3}.10.\frac{7}{35}.0,75\)
\(=\frac{1}{2}.\frac{4}{3}.10.\frac{7}{35}.\frac{3}{4}\)
\(=\left(\frac{1}{2}.10\right)\times\left(\frac{4}{3}.\frac{3}{4}\right).\frac{7}{35}\)
\(=5.1.\frac{7}{35}\)
\(=\frac{35}{35}=1\)
~ Hok tốt ~
10 tháng 5 2019
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{40}-\frac{1}{40}\right)-\frac{1}{43}\)
\(=1-\frac{1}{43}=\frac{42}{43}\)
~ Hok tốt ~
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{40.43}\)
\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\right)\)
\(=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\right)\)
\(=\frac{1}{3}\left(1-\frac{1}{43}\right)\)
\(=\frac{1}{3}.\frac{42}{43}=\frac{14}{43}\)
A= \(\frac{1}{1x4}\)+\(\frac{1}{4x7}\)+\(\frac{1}{7x10}\)+......+\(\frac{1}{39x43}\)
3A=\(\frac{3}{1x4}\)+\(\frac{3}{4x7}\)+\(\frac{3}{7x10}\)+....+ \(\frac{3}{39x43}\)
3A=\(\frac{1}{1}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{10}\)+....+\(\frac{1}{39}\)-\(\frac{1}{43}\)
3A=\(\frac{1}{1}\)-\(\frac{1}{43}\)
3A=\(\frac{42}{43}\)
A=\(\frac{42}{129}\)=\(\frac{14}{43}\)
k mình nha!!!!!!!!!