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TA
1
2 tháng 8 2023
\(A=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{199}+\dfrac{1}{120}\left(a\right)\)
\(\Rightarrow A=\left(\dfrac{1}{101}+\dfrac{1}{102}+...\dfrac{1}{125}\right)+\left(\dfrac{1}{126}+\dfrac{1}{127}+...\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...\dfrac{1}{175}\right)+\left(\dfrac{1}{176}+\dfrac{1}{177}+...\dfrac{1}{200}\right)\)
\(\Rightarrow A>25.\dfrac{1}{125}+25.\dfrac{1}{150}+25.\dfrac{1}{175}+25.\dfrac{1}{200}\)
\(\Rightarrow A>\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\)
\(\Rightarrow A>\dfrac{168+140+120+105}{840}=\dfrac{533}{840}>\dfrac{5}{8}\left(\dfrac{533}{840}>\dfrac{525}{840}\right)\)
\(\Rightarrow A>\dfrac{5}{8}\left(1\right)\)
\(\left(a\right)\Rightarrow A=\left(\dfrac{1}{101}+...\dfrac{1}{120}\right)+\left(\dfrac{1}{121}+...\dfrac{1}{140}\right)+\left(\dfrac{1}{141}+...\dfrac{1}{160}\right)+\left(\dfrac{1}{161}+...\dfrac{1}{180}\right)+\left(\dfrac{1}{181}+...\dfrac{1}{200}\right)\)
\(\Rightarrow A< 20.\dfrac{1}{100}+20.\dfrac{1}{120}+20.\dfrac{1}{140}+20.\dfrac{1}{160}+20.\dfrac{1}{180}\)
\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}\)
\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{504+420+360+315+280}{2520}=\dfrac{1879}{2520}< \dfrac{3}{4}\left(\dfrac{1879}{2520}< \dfrac{1890}{2520}\right)\)
\(\Rightarrow A< \dfrac{3}{4}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\dfrac{5}{8}< A< \dfrac{3}{4}\left(dpcm\right)\)
HA
0
VN
1
15 tháng 8 2017
Ta có : \(\frac{1}{101}\) > \(\frac{1}{150}\)
\(\frac{1}{102}\) > \(\frac{1}{150}\)
.....................................................
\(\frac{1}{149}\) > \(\frac{1}{150}\)
=> \(\frac{1}{101}\) + \(\frac{1}{102}\) + .......... + \(\frac{1}{150}\) > \(\frac{1}{150}\) + \(\frac{1}{150}\) + .......... + \(\frac{1}{150}\)( có 50 p/s ) = \(\frac{1}{150}\) . 50 = \(\frac{1}{3}\)(1)
Ta lại có : \(\frac{1}{151}\) > \(\frac{1}{200}\)
\(\frac{1}{152}\) > \(\frac{1}{200}\)
............................................
\(\frac{1}{199}\)> \(\frac{1}{200}\)
=> \(\frac{1}{151}\) + \(\frac{1}{152}\) + .................. + \(\frac{1}{200}\) > \(\frac{1}{200}\)+ \(\frac{1}{200}\) + ...................+ \(\frac{1}{200}\)(có 50 p/ )=\(\frac{1}{200}\) . 50 = \(\frac{1}{4}\)(2)
Từ (1) và (2)
=> \(\frac{1}{101}\)+ \(\frac{1}{102}\) + \(\frac{1}{103}\) + ...................+ \(\frac{1}{200}\)> \(\frac{1}{3}\) + \(\frac{1}{4}\) = \(\frac{4}{12}\) + \(\frac{3}{12}\) = \(\frac{7}{12}\)
Vậy A > \(\frac{7}{12}\)
27 tháng 9 2019
\(A=\frac{1}{101^2}+\frac{1}{102^2}+\frac{1}{103^2}+\frac{1}{104^2}+\frac{1}{105^2}\)
\(A< \frac{1}{100\cdot101}+\frac{1}{101\cdot102}+\frac{1}{102\cdot103}+\frac{1}{103\cdot104}+\frac{1}{104\cdot105}\)
\(=\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+\frac{1}{103}-\frac{1}{104}+\frac{1}{104}-\frac{1}{105}\)
\(=\frac{1}{100}-\frac{1}{105}=\frac{1}{2100}=\frac{1}{2^2\cdot3\cdot5^2\cdot7}=B\)
Vậy \(A< B\)
7 tháng 7 2021
Ta xét: \(\dfrac{1}{100} + \dfrac{1}{101} + \dfrac{1}{102}...+ \dfrac{1}{200}\)
\(\dfrac{1}{100} > \dfrac{1}{200}\)
\(\dfrac{1}{101}>\dfrac{1}{200}\)
.
.
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\(\dfrac{1}{199}>\dfrac{1}{200}\)
\(\Rightarrow\)\(\dfrac{1}{100} + \dfrac{1}{101} + \dfrac{1}{102} +...+\dfrac{1}{200}\)(có 101 phân số) > \(100.\dfrac{1}{200} = \dfrac{1}{2}\)
AH
Akai Haruma
Giáo viên
7 tháng 7 2021
Lời giải:
\(\frac{1}{100}+\frac{1}{101}+...+\frac{1}{200}>\frac{1}{200}+\frac{1}{200}+....+\frac{1}{200}=\frac{101}{200}>\frac{100}{200}=0,5>0,499\)