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\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}\)
= \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}\)
\(=\dfrac{1}{2}-\dfrac{1}{17}\)
\(=\dfrac{15}{34}\)
Vì \(\dfrac{15}{34}< \dfrac{1}{2}=>\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot27}< \dfrac{1}{2}\)
Cau a) 1/1.2 +1/2.3 +1/3.4+...+1/99.100= 1/1-1/2+1/2-1/3+...+1/99-1/100
=1/1-1/100=99/100
99/100<1 thì 1/1.2 +1/2.3+1/3.4+...+1/99.100<1
Câu b): Ta có: 1/2^2<1/1.2
1/3^2<1/2.3
...............(so sánh như vậy với các số khác)
1/2016^2<1/2015.2016
Áp dụng của câu a ta thêm vào sau về thành: 1/1.2+1/2.3+1/3.4+...+1/2015.2016
=1/1-1/2+1/2-1/3+1/3-1/4+...+1/2015-1/2016
=1/1-1/2016
=2015/2016<1
Ma :1/2^2+1/3^2+1/4^2+...+1/2016^2<1/1.1+1/2.3+1/3.4+...+1/2015.2016
Nen:1/1^2+1/3^2+1/4^2+...+1/2016^2<1
\(a,A=\frac{3}{2}+\frac{3}{6}+\frac{3}{12}+\frac{3}{20}+...+\frac{3}{90}\)
\(A=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(A=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=3.\left(1-\frac{1}{10}\right)\)
\(A=3.\frac{9}{10}=\frac{27}{10}\)
\(b,B=\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+\frac{2}{14.17}\)
\(B.\frac{3}{2}=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\)
\(B.\frac{3}{2}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)
\(B.\frac{3}{2}=\frac{1}{2}-\frac{1}{17}\)
\(B=\frac{15}{34}:\frac{3}{2}=\frac{5}{17}\)
\(\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+...+\frac{2}{14.17}=2.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{14.17}\right)\)
\(=\frac{2}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{14}-\frac{1}{17}\right)\)
\(=\frac{2}{3}.\left(\frac{1}{2}-\frac{1}{17}\right)=\frac{2}{3}.\frac{15}{34}=\frac{5}{17}\)
\(\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+...+\frac{2}{14.17}\)
\(=\frac{2}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{14.17}\right)\)
\(=\frac{2}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{14}-\frac{1}{17}\right)\)
\(=\frac{2}{3}.\left(\frac{1}{2}-\frac{1}{17}\right)\)
\(=\frac{2}{3}.\left(\frac{17}{34}-\frac{2}{34}\right)\)
\(=\frac{2}{3}.\frac{15}{34}=\frac{5}{17}\)
M = 4/2.5 + 4/5.8 + 4/8.11 + 4/11.14 + 4/14.17 + 4/17.20
M= 4/3 . (1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17+1/17-1/20)
M= 4/3 . (1/2 - 1/20)
M= 4/3 . (10/20 - 1/20)
M= 4/3 . 9/20
M= 3/5
k nha
khi ko mún tích thì tích 1 tích
khi mún tích thì tích 50 tích
S = 1/2.5 +1/5.8 +1/8.11+1/11.14+1/14.17+1/17.20
S=1/3.(1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17+1/17-1/20)
S=1/3.(1/2-1/20)
S=1/3.(10/20-1/20)
S=1/3.9/20
S= 3/20
k nha
\(B=\frac{9}{8\cdot11}+\frac{9}{11\cdot14}+...+\frac{9}{197\cdot200}\)
\(=3\left(\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{197\cdot200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)
\(=3\left(\frac{1}{8}-\frac{1}{200}\right)\)
\(=3\left(\frac{24}{200}-\frac{1}{200}\right)\)
\(=3\cdot\frac{23}{200}\)
đúng
a/
\(\dfrac{3A}{2}=\dfrac{8-5}{5.8}+\dfrac{11-8}{8.11}+\dfrac{14-11}{11.14}+...+\dfrac{26-29}{26.29}=\)
\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{26}-\dfrac{1}{29}=\)
\(=\dfrac{1}{5}-\dfrac{1}{29}\)
Từ đó tính ra A bạn tự tính nốt nhé
b/
\(B=-137.136+137.417-136.417+136.137=\)
\(=417.\left(137-136\right)=417\)
B) -137. ( 136 - 417 ) - 136. ( 417 - 137 )
= -137. (-281) - 136. 280
= 38497 - 38080
= 417