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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 \(\dfrac{15}{34}< \dfrac{1}{2}=>\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot27}< \dfrac{1}{2}\)
\(\frac{3}{2.5}\)+\(\frac{3}{5.8}\)+\(\frac{3}{8.11}\)+\(\frac{3}{11.14}\)+\(\frac{3}{14.17}\)
=\(\frac{1}{2}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{8}\)+......+\(\frac{1}{14}\)-\(\frac{1}{17}\)
=\(\frac{1}{2}\)-\(\frac{1}{17}\)
=\(\frac{15}{34}\)
Đặt \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\)
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)
\(A=\frac{1}{2}-\frac{1}{17}\)
\(A=\frac{15}{34}\)
= \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)= \(\frac{1}{2}-\frac{1}{17}\)=\(\frac{15}{34}\)
x.\(\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\right)=-1\frac{3}{5}\)
x.\(\left(\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+\frac{14-11}{11.14}+\frac{17-14}{14.17}\right)=\frac{-8}{5}\)
x.\(\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)=\frac{-8}{5}\)
x.\(\left(\frac{1}{2}-\frac{1}{17}\right)=\frac{-8}{5}\)
x.\(\left(\frac{17}{34}-\frac{2}{34}\right)=\frac{-8}{5}\)
x.\(\frac{15}{34}=\frac{-8}{5}\)
x\(=\frac{-8}{5}:\frac{15}{34}\)
x\(=\frac{-8}{5}.\frac{34}{15}\)
x\(=\frac{-272}{75}\)
Vậy x\(=\frac{-272}{75}\)
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
=1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14
=1/2-1/14
=7/14-1/14=6/14=3/7
3x/2.5 + 3x/5.8 + 3x/8.11 + 3x/11.14 = 1/21
=> x . ( 3/2.5 + 3/5.8 + 3/8.11 + 3/11.14 ) = 1/21
=> x . ( 1/2.5 + 1/5.8 + 1/8.11 + 1/11.14 ) = 1/21
x . ( 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14 ) = 1/21
x . ( 1/2 - 1/14 ) = 1/21
x . 3/7 = 1/21
x = 1/21 : 3/7
=> x = 1/9
\(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)
<=> \(x\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right)=\frac{1}{21}\)
<=> \(x\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)
<=> \(x\left(\frac{1}{2}-\frac{1}{14}\right)=\frac{1}{21}\)
<=> \(x\cdot\frac{3}{7}=\frac{1}{21}\)
<=> \(x=\frac{1}{9}\)
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
\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}\\ =\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\\ =\frac{1}{2}-\frac{1}{17}\\ =\frac{17}{34}-\frac{2}{34}=\frac{15}{34}\)
thanks