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![](https://rs.olm.vn/images/avt/0.png?1311)
= 1 -1/4 +1/4 -1/7 +1/7 -1/10+...+1/40 -1/43 +1/43 -1/46
= 1 -1/46
= 45/46
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}\)
\(=\frac{45}{46}\)
_Chúc bạn học tốt_
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\) < 1
\(S=3\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{40.43}+\frac{1}{43.46}\right)\)
\(S=3.\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\right)\)
\(\Rightarrow S=1-\frac{1}{46}\Rightarrow S< 1\left(đpcm\right)\)
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
= \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
= \(1-\frac{1}{46}< 1\)
\(\Rightarrow S< 1\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}< 1\)
Vậy \(S< 1\)
Chúc bạn học tốt !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(=1-\left(\frac{1}{4}-\frac{1}{4}\right)-\left(\frac{1}{7}-\frac{1}{7}\right)-\left(\frac{1}{10}-\frac{1}{10}\right)-...-\left(\frac{1}{40}-\frac{1}{40}\right)-\frac{1}{43}\)
\(=1-\frac{1}{43}=\frac{42}{43}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+....+\frac{3}{40\cdot43}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{40}-\frac{1}{43}\)
\(=1-\frac{1}{43}=\frac{42}{43}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{43.46}< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
7/3 - 2/3 . x = 7/4
2/3 . x = 7/3 - 7/4
2/3 . x = 28/12 - 21/ 12
2/3 . x = 7/12
x = 7/12 : 2/3
x = 7/12 . 3/2
x = 21/ 24
x = 7/8
vậy x = 7/8
\(\frac{7}{3}-\frac{2}{3}x=-1\frac{3}{4}\)
\(\frac{7}{3}-\frac{2}{3}x=-\frac{7}{4}\)
\(\frac{2}{3}x=\frac{7}{3}-\left(-\frac{7}{4}\right)\)
\(\frac{2}{3}x=\frac{7}{3}+\frac{7}{4}\)
\(\frac{2}{3}x=\frac{49}{12}\)
\(x=\frac{49}{12}:\frac{2}{3}\)
\(x=\frac{49}{12}\cdot\frac{3}{2}\)
\(x=\frac{49}{8}\)
nếu thấy đúng thì k cho mh nhahoàng mộc miên123___
\(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}+\frac{3}{43.46}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)
\(S=1-\frac{1}{46}< 1\)
Chứng tỏ S < 1
Ủng hộ mk nha ^_^
S = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{43.46}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{43}-\frac{1}{46}\)
\(=1-\frac{1}{46}=\frac{45}{46}< 1\)