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Đặt \(A=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\)
\(A=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)
\(A=\frac{1}{2}-\frac{1}{17}\)
\(A=\frac{15}{34}\)
= \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)= \(\frac{1}{2}-\frac{1}{17}\)=\(\frac{15}{34}\)
\(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{21.24}\)
\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{21}-\frac{1}{24}\)
\(=\frac{1}{5}-\frac{1}{24}\)
\(=\frac{19}{120}\)
Ta có :
\(A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2009.2012}+\frac{3}{2012.2015}\)
\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2009}-\frac{1}{2012}+\frac{1}{2012}-\frac{1}{2015}\)
\(A=\frac{1}{5}-\frac{1}{2015}\)
\(A=\frac{402}{2015}\)
Vậy \(A=\frac{402}{2015}\)
Chúc bạn học tốt ~
\(\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}\)
=\(3\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\right)\)
= \(3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)
= \(3\left(\frac{1}{2}-\frac{1}{17}\right)\)
=\(\frac{45}{34}\)
\(3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\right)\)
=3(3/2.5+3/5.8+3/8.11+3/11.14+3/14.17)
=3(1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17)
=3(1/2-1/17)
=45/34
cô Nhung ơi k đúng cho con đi cô pls
\(\frac{3}{2.5}\)+\(\frac{3}{5.8}\)+\(\frac{3}{8.11}\)+\(\frac{3}{11.14}\)+\(\frac{3}{14.17}\)
=\(\frac{1}{2}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{8}\)+......+\(\frac{1}{14}\)-\(\frac{1}{17}\)
=\(\frac{1}{2}\)-\(\frac{1}{17}\)
=\(\frac{15}{34}\)
Dat A=3/5.8 +3/8.11 +.........+3/32.35
A=1/5-1/8+1/8-1/11+1/11-1/14+.............+1/32-1/35
A=1/5-1/35
A=6/35
=>x+6/35=-29/35
=>x=-29/35-6/35
=>x=-1
\(H=\frac{3^2}{2.5}+\frac{3^2}{5.8}+\frac{3^2}{8.11}+\frac{3^2}{11.14}+...+\frac{3^2}{197.200}=3.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{197.200}\right)=3.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{197}-\frac{1}{200}\right)=3\cdot\left(\frac{1}{2}-\frac{1}{200}\right)==\frac{297}{200}\)
\(a,A=\frac{3}{2}+\frac{3}{6}+\frac{3}{12}+\frac{3}{20}+...+\frac{3}{90}\)
\(A=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right)\)
\(A=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=3.\left(1-\frac{1}{10}\right)\)
\(A=3.\frac{9}{10}=\frac{27}{10}\)
\(b,B=\frac{2}{2.5}+\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+\frac{2}{14.17}\)
\(B.\frac{3}{2}=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}\)
\(B.\frac{3}{2}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)
\(B.\frac{3}{2}=\frac{1}{2}-\frac{1}{17}\)
\(B=\frac{15}{34}:\frac{3}{2}=\frac{5}{17}\)
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Rightarrow3\left(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}\right)=3.\frac{101}{1540}\)
\(\Rightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
\(\Rightarrow x+3=308\)
\(\Rightarrow x=305\)
\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\) (x khác 0; khác -3)
\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
<=>\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
<=>\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
<=>\(\frac{1}{x+3}=\frac{1}{308}\)
=>x+3=308
<=>x=305 (nhận)
Vậy x=305
\(B=\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{2018.2021}\)
\(B=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{2018}-\frac{1}{2021}\)
\(B=\frac{1}{5}-\frac{1}{2021}\)
\(B=\frac{2016}{10105}\)