Giải thích các bước giải:

Bài `4.`
`a)`
`A=((1,5+1-0,75)/(2,5+5/3-1,25)+(0,375-0,3+3/11+3/12)/(-0,625+0,5-5/11-5/12)). 1890/2005+115`
`A=((15/10+3/3-75/100)/(25/10+5/3-125/100)+(375/1000-3/10+3/11+3/12)/(-625/1000+5/10-5/11-5/12)) . 378/401+115`
`A=((3/2+3/3-3/4)/(5/2+5/3-5/4)+(3/8-3/10+3/11+3/12)/(-5/8+5/10-5/11-5/12)) . 378/401+115`
`A=((3.(1/2+1/3-1/4))/(5.(1/2+1/3-1/4))+(3.(1/8-1/10+1/11+1/12))/(-(5/8-5/10+5/11+5/12))). 378/401+115`
`A=((3.(1/2+1/3-1/4))/(5.(1/2+1/3-1/4))-(3.(1/8-1/10+1/11+1/12))/(5.(1/8-1/10+1/11+1/12))). 378/401+115`
`A=(3/5-3/5). 378/401 +115`
`A=0. 378/401+115`
`A=0+115`
`A=115`
Vậy `A=115`
`b)`
`B=1/3+1/3^2+1/3^3+1/3^4+...+1/3^2004+1/3^2005`
`3 . B=3.(1/3+1/3^2+1/3^3+1/3^4+...+1/3^2004+1/3^2005)`
`3B=1+1/3+1/3^2+1/3^3+...+1/3^2003+1/3^2004`
`3B-B=(1+1/3+1/3^2+1/3^3+...+1/3^2003+1/3^2004)-(1/3+1/3^2+1/3^3+1/3^4+...+1/3^2004+1/3^2005)`
`B.(3-1)=1+1/3+1/3^2+1/3^3+...+1/3^2003+1/3^2004-1/3-1/3^2-1/3^3-1/3^4-...-1/3^2004-1/3^2005`
`2B=1+(1/3-1/3)+(1/3^2-1/3^2)+(1/3^3-1/3^3)+...+(1/3^2003-1/3^2003)+(1/3^2004-1/3^2004)-1/3^2005`
`2B=1-1/3^2005`
`B=(1-1/3^2005)/2`
`B=1/2-(1/3^2005)/2<1/2`
`=>B<1/2(text{ĐPCM})`
Vậy `B=1/3+1/3^2+1/3^3+1/3^4+...+1/3^2004+1/3^2005<1/2`

Đáp án:

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Giải thích các bước giải:

$\left(\dfrac{1,5+1-0,75}{2,5+\dfrac53-1,25}+\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac5{11}-\dfrac5{12}}\right):\dfrac{1890}{2005}+115\\=\left(\dfrac{\dfrac32+\dfrac33-\dfrac34}{\dfrac52+\dfrac53-\dfrac54}+\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{\dfrac{-5}{8}-\dfrac{-5}{10}+\dfrac{-5}{11}+\dfrac{-5}{12}}\right):\dfrac{396}{401}+115\\=\left(\dfrac{3.\left(\dfrac12+\dfrac13-\dfrac14\right)}{5.\left(\dfrac12+\dfrac13-\dfrac14\right)}+\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}\right):\dfrac{396}{401}+115\\=\left(\dfrac35+\dfrac{-3}5\right):\dfrac{396}{401}+115\\=0:\dfrac{396}{401}+115\\=115\\B=\dfrac13+\dfrac{1}{3^2}+\dfrac{1}{3^3}\ +\,.\!.\!.+\ \dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\\\Rightarrow 3B=3.\left(\dfrac13+\dfrac{1}{3^2}+\dfrac{1}{3^3}\ +\,.\!.\!.+\ \dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\right)\\\Rightarrow 3B=1+\dfrac13+\dfrac{1}{3^2}\ +\,.\!.\!.+\ \dfrac{1}{3^{2003}}+\dfrac{1}{3^{2004}}\\\Rightarrow 3B -B=\left(1+\dfrac13+\dfrac{1}{3^2}\ +\,.\!.\!.+\ \dfrac{1}{3^{2003}}+\dfrac{1}{3^{2004}}\right)-\left(\dfrac13+\dfrac{1}{3^2}+\dfrac{1}{3^3}\ +\,.\!.\!.+\ \dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\right)\\\Rightarrow 2B=1-\dfrac{1}{3^{2005}}\\\Rightarrow B=\dfrac{1-\dfrac{1}{3^{2005}}}{2}\\\Rightarrow B=\dfrac12-\dfrac{1}{2.3^{2005}}<\dfrac12\\\Rightarrow B<\dfrac12$