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`5/13 xx 2/7 +12/7 xx 5/13 - 5/13`
`= 5/13 xx 2/7 +12/7 xx 5/13 - 5/13xx1`
`= 5/13 xx (2/7 + 12/7 - 1)`
`= 5/13 xx ( 14/7 -1 )`
`= 5/13 xx ( 2-1)`
`= 5/13 xx 1`
`= 5/13`
\(\dfrac{5}{13}x\dfrac{2}{7}+\dfrac{12}{7}x\dfrac{5}{13}-\dfrac{5}{13}=\dfrac{5}{3}x\left(\dfrac{2}{7}+\dfrac{12}{7}-1\right)=\dfrac{5}{3}x\left(2-1\right)=\dfrac{5}{3}\)
Ta có : B = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\)
=> 3B = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)
Khi đó 3B - B = \(\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2005}}\right)\)
=> 2B = \(1-\frac{1}{3^{2005}}\)
=> B = \(\frac{1}{2}-\frac{1}{3^{2005}.2}< \frac{1}{2}\left(\text{ĐPCM}\right)\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+........+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)
\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+........+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
\(\Rightarrow3B-B=1-\frac{1}{3^{2005}}\)
\(\Rightarrow2B=1-\frac{1}{3^{2005}}\)\(\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}\)
Vì \(1-\frac{1}{3^{2005}}< 1\)\(\Rightarrow\frac{1-\frac{1}{3^{2005}}}{2}< \frac{1}{2}\)
hay \(B< \frac{1}{2}\)( đpcm )
13 x 2/ 3=
26/3
75,3333333333