bài 2Tính
\(\frac{3^3}{_{6.11}}+\frac{3^3}{11.16}+....+\frac{3^3}{91.96}\)
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NT
1
DP
5 tháng 7 2017
\( A=\frac{3^3}{6\cdot11}+\frac{3^3}{11\cdot16}+\frac{3^3}{16\cdot21}+....+\frac{3^3}{91\cdot96}\)
\(A=\frac{3^3}{5}\cdot\left(\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{91}-\frac{1}{96}\right)\)
\(A=\frac{27}{5}\cdot\left(\frac{1}{6}-\frac{1}{96}\right)\)
\(A=\frac{27}{5}\cdot\frac{5}{32}=\frac{27}{32}\)
H
2
29 tháng 7 2016
\(\Leftrightarrow B=\frac{3}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\right)\)
\(\Leftrightarrow B=\frac{3}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(\Leftrightarrow B=\frac{3}{5}.\frac{100}{101}\)
\(\Leftrightarrow B=\frac{60}{101}\)
H
0
NH
3
12 tháng 8 2016
\(.S=3.\left(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{96.101}\right)\)
\(\Rightarrow S=3.\frac{1}{5}\left(\frac{1}{1}-\frac{1}{6}+...+\frac{1}{96}-\frac{1}{101}\right)\)
\(\Rightarrow S=\frac{3}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(\Rightarrow S=\frac{3}{5}.\left(\frac{100}{101}\right)\)
\(S=\frac{60}{101}\)
NT
2
HT
11 tháng 8 2016
\(\frac{3}{1.6}+\frac{3}{6.11}+\frac{3}{11.16}+...+\frac{3}{96.101}\)
\(=3.\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)\)
\(=\frac{3}{5}.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)\)
\(=\frac{3}{5}.\left(1-\frac{1}{101}\right)\)
\(=\frac{3}{5}.\frac{100}{101}\)
\(=\frac{60}{101}\)
28 tháng 3 2017
a,1/1-1/4+1/4-1/7+...+1/2008-1/2011
=(1-1/2011)+(-1/4+1/4)+...+(-1/2008+1/2008)
=1-1/2011+0+...+0
=1-1/2011
=2010/2011
9 tháng 1 2020
\(b\)) \(Q=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)
\(=5.\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5.\left(1-\frac{1}{31}\right)=\frac{150}{31}\)
9 tháng 1 2020
\(a\)) Mình giải theo cách khác:
Chú ý rằng : \(\frac{3}{2.5}=\frac{1}{2}-\frac{1}{5};\frac{3}{5.8}=\frac{1}{5}-\frac{1}{8};\frac{3}{8.11}=\frac{1}{8}-\frac{1}{11};...;\frac{3}{17.20}=\frac{1}{17}-\frac{1}{20}\)
Do đó: \(P=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)
12 tháng 4 2018
Ta có :
\(S=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}\)
\(S=5\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}\right)\)
\(S=5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}\right)\)
\(S=5\left(1-\frac{1}{26}\right)\)
\(S=5.\frac{25}{26}\)
\(S=\frac{125}{26}\)
Vậy \(S=\frac{125}{26}\)
Chúc bạn học tốt ~
12 tháng 6 2017
Câu 1:
Giả sử \(\frac{3}{5}< \frac{3+m}{5+m}\)
=) \(3.\left(5+m\right)< 5.\left(3+m\right)\)
=) \(15+3m< 15+5m\) ( Đúng vì \(15=15\)và \(3m< 5m\)) =) Điều giả sử đúng
=) \(\frac{3}{5}< \frac{3+m}{5+m}\)
* Từ điều trên ta suy ra : Nếu \(\frac{a}{b}< 1\)=) \(\frac{a}{b}< \frac{a+m}{b+m}\)
Và nếu \(\frac{a}{b}>1\)=) \(\frac{a}{b}>\frac{a+m}{b+m}\)
Câu 2 :
= \(5.\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right)\)
= \(5.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
= \(5.\left(\frac{1}{1}-\frac{1}{31}\right)\)= \(5.\frac{30}{31}=\frac{150}{31}\)
Q
12 tháng 6 2017
=> Với mọi số tự nhiên m ( như m\(\ne\)0 ) thì \(\frac{3}{5}< \frac{3+m}{5+m}\)
\(\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}\)
\(=5\left(\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{26.31}\right)\)
\(=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5\left(1-\frac{1}{31}\right)\)
\(=5.\frac{30}{31}\)
\(=\frac{150}{31}\)
\(\frac{3^3}{6.11}+\frac{3^3}{11.16}+.......+\frac{3^3}{91.96}\)
\(=\frac{3^3}{5}.\left(\frac{5}{6.11}+\frac{5}{11.16}+.....+\frac{5}{91.96}\right)\)
\(=\frac{27}{5}.\left(\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+.....+\frac{1}{91}-\frac{1}{96}\right)\)
\(=\frac{27}{5}.\left(\frac{1}{6}-\frac{1}{96}\right)\)
\(=\frac{27}{5}.\frac{5}{32}\)
\(=\frac{27}{32}\)
\(\frac{3^3}{6\cdot11}+\frac{3^3}{11\cdot16}+...+\frac{3^3}{91\cdot96}\)
\(=\frac{3^3}{5}\cdot\left(\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{91}-\frac{1}{96}\right)\)
\(=\frac{27}{5}\cdot\left(\frac{1}{6}-\frac{1}{96}\right)\)
\(=\frac{27}{5}\cdot\frac{5}{32}=\frac{27}{32}\)
Vậy \(\frac{3^3}{6\cdot11}+\frac{3^3}{11\cdot16}+...+\frac{3^3}{91\cdot96}=\frac{27}{32}\)