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\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{899}{900}\)
\(A=\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{29\cdot31}{30\cdot30}\)
\(A=\dfrac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot31}{\left(2\cdot3\cdot4\cdot...\cdot30\right)^2}\)
\(A=\dfrac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot31}{\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot30}\)
\(A=\dfrac{1\cdot31}{30}=\dfrac{31}{30}\)
Ta có : \(\dfrac{1}{101}>\dfrac{1}{300}\)
...
\(\dfrac{1}{299}>\dfrac{1}{300}\)
Do đó :
\(\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{300}>\dfrac{1}{300}+\dfrac{1}{300}..+\dfrac{1}{300}\)
\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{300}>\dfrac{200}{300}=\dfrac{2}{3}\)
Vậy...
a)
Ta thấy:
\(\dfrac{1}{6}< \dfrac{1}{5}\)
\(\dfrac{1}{7}< \dfrac{1}{5}\)
\(\dfrac{1}{8}< \dfrac{1}{5}\)
\(\dfrac{1}{9}< \dfrac{1}{5}\)
\(\dfrac{1}{11}< \dfrac{1}{10}\)
\(\dfrac{1}{12}< \dfrac{1}{10}\)
\(\dfrac{1}{13}< \dfrac{1}{10}\)
...
\(\dfrac{1}{17}< \dfrac{1}{10}\)
\(\Rightarrow\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 5\cdot\dfrac{1}{5}+8\cdot\dfrac{1}{10}=1+\dfrac{4}{5}=\dfrac{9}{5}< 2\)
Vậy \(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}< 2\)
b)
Ta thấy:
\(\dfrac{1}{101}>\dfrac{1}{300}\)
\(\dfrac{1}{102}>\dfrac{1}{300}\)
\(\dfrac{1}{103}>\dfrac{1}{300}\)
...
\(\dfrac{1}{299}>\dfrac{1}{300}\)
\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{300}>200\cdot\dfrac{1}{300}=\dfrac{2}{3}\)
Vậy \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{300}>\dfrac{2}{3}\)
\(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{10^2}\)
\(\Leftrightarrow D=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{10.10}\)
\(\Leftrightarrow D< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(\Leftrightarrow D< \dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{10-9}{9.10}\)
\(\Leftrightarrow D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Leftrightarrow D< 1-\dfrac{1}{10}\)
\(\Leftrightarrow D< \dfrac{9}{10}< \dfrac{10}{10}=1\)
\(\Leftrightarrow D< 1\left(đpcm\right)\)
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{199}-\dfrac{1}{200}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{199}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+..+\dfrac{1}{200}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{200}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)\)
\(=\dfrac{1}{101}+...+\dfrac{1}{199}+\dfrac{1}{200}\)
\(B=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{64}\\ B=1+\dfrac{1}{2}+\left(\dfrac{1}{3}+\dfrac{1}{4}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right)+\left(\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{16}\right)+\left(\dfrac{1}{17}+\dfrac{1}{18}+...+\dfrac{1}{32}\right)+\left(\dfrac{1}{33}+\dfrac{1}{34}+...+\dfrac{1}{64}\right)\\ B>1+\dfrac{1}{2}+\left(\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}+\dfrac{1}{8}\right)+\left(\dfrac{1}{16}+\dfrac{1}{16}+...+\dfrac{1}{16}\right)+\left(\dfrac{1}{32}+\dfrac{1}{32}+...+\dfrac{1}{32}\right)+\left(\dfrac{1}{64}+\dfrac{1}{64}+...+\dfrac{1}{64}\right)\\ B>1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\\ B>4\)
a. ta có
1/101 > 1/150
1/102> 1/150
...>1/150
1/150 = 1/150
=> 1/101 + 1/102 + .... + 1/150 > 1/150 +1/150+....+1/150(50 số hạng )= 1/3
ta có
1/151 >1/200
1/152 > 1/200
..>1/200
1/200 = 1/200
=> 1/151 + 1/152+....+1/200 > 1/200+1/200+ ...+1/200( 50 số hạng) = 1/4
==> 1/101 + 1/102+....+1/200 > 1/3 +1/4
==> A > 7/12
b, A =(1/101+1/102+....+1/150)+(1/151+1/152+.....+1/200)
A>1/150.50+1/200.50=1/3+1/4=7/12
b tách A thành bốn nhóm rồi cũng làm như trên,ta có
A>25/125+25/150+25/175+25/200=(1/5+1/6+1/7)+1/8
=107/210+1/8>1/2+1/8=5/8
Ta có:
\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{299}+\dfrac{1}{300}>\dfrac{1}{300}.200=\dfrac{200}{300}=\dfrac{2}{3}\)
\(\Rightarrow\) biểu thức trên lớn hơn \(\dfrac{2}{3}\).