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\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{199}-\dfrac{1}{200}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{199}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+..+\dfrac{1}{200}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{200}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{199}+\dfrac{1}{200}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)\)
\(=\dfrac{1}{101}+...+\dfrac{1}{199}+\dfrac{1}{200}\)
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:
\(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{150}>\dfrac{1}{150}+\dfrac{1}{150}+\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\) (có 50 số hạng)
⇔ \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{150}>\dfrac{1}{3}\) \(\left(1\right)\)
\(\dfrac{1}{151}+\dfrac{1}{152}+\dfrac{1}{153}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}\) (có 50 số hạng)
⇔ \(\dfrac{1}{151}+\dfrac{1}{152}+\dfrac{1}{153}+...+\dfrac{1}{200}>\dfrac{1}{4}\) \(\left(2\right)\)
Từ (1) và (2), cộng vế theo vế. Ta được:
\(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{150}+\dfrac{1}{151}+\dfrac{1}{152}+\dfrac{1}{153}+...+\dfrac{1}{200}>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\)
⇒ \(ĐPCM\)
\(B=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+...\dfrac{1}{200}\right)>\dfrac{1}{150}+..\dfrac{1}{150}+\dfrac{1}{200}+..+200=\dfrac{50}{150}+\dfrac{50}{200}=\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{4}{12}+\dfrac{3}{12}=\dfrac{7}{12}\)Vậy ... (ta có điều phải chứng minh )
Ta có :\(\dfrac{1}{20}>\dfrac{1}{200}\)
...
\(\dfrac{1}{199}>\dfrac{1}{200}\)
Do đó : \(\dfrac{1}{20}+\dfrac{1}{21}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+..+\dfrac{1}{200}=\dfrac{181}{200}>\dfrac{180}{200}=\dfrac{9}{10}\)Vậy ...
\(Tacó\)
\(\dfrac{1}{101}>\dfrac{1}{200}\)
\(\dfrac{1}{102}>\dfrac{1}{200}\)
...
\(\dfrac{1}{999}>\dfrac{1}{200}\)
Do đó :\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{999}+\dfrac{1}{200}>\dfrac{1}{200}+...+\dfrac{1}{200}=100.\dfrac{1}{200}=\dfrac{100}{200}=\dfrac{1}{2}\)
Ta lại có:
\(\dfrac{1}{102}< \dfrac{1}{101}\)
\(\dfrac{1}{103}< \dfrac{1}{101}\)
...
\(\dfrac{1}{200}< \dfrac{1}{101}\)
Do đó : \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}< \dfrac{1}{101}+\dfrac{1}{101}+...+\dfrac{1}{101}=\dfrac{1}{101}.100=\dfrac{100}{101}< 1\)Vậy ...( theo tớ , cậu nên đánh dấu (1) và (2) rồi suy ra ) .. khẳng định trên , học tốt
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 \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{300}>\dfrac{2}{3}\)
a: A>1/150*50+1/200*50=1/3+1/4=7/12
b: A>7/12
7/12>5/8
=>A>5/8
Ta có:
\(\dfrac{1}{101}>\dfrac{1}{150}\)
\(\dfrac{1}{102}>\dfrac{1}{150}\)
....
\(\dfrac{1}{150}=\dfrac{1}{150}\)
=>\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}>\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\)(50 số)=\(\dfrac{1}{3}\)
Ta có:
\(\dfrac{1}{152}>\dfrac{1}{200}\)
\(\dfrac{1}{153}>\dfrac{1}{200}\)
....
\(\dfrac{1}{200}=\dfrac{1}{200}\)
=>\(\dfrac{1}{151}+\dfrac{1}{153}+...+\dfrac{1}{120}>\dfrac{1}{120}+\dfrac{1}{120}+...+\dfrac{1}{120}\)(50 số)=\(\dfrac{1}{4}\)
=>\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}>\dfrac{1}{3}+\dfrac{1}{4}\)
=> \(A>\dfrac{7}{12}\)
Cảm ơn bạn.