a) Ta có: \(\frac{2010}{2009}=1+\frac{1}{2009}\)(1)

            \(\frac{2011}{2010}=1+\frac{1}{2010}\)(2)

Từ (1) và (2)

    Mà: \(\frac{1}{2009}>\frac{1}{2010}\)

       \(\Rightarrow\frac{2010}{2009}>\frac{2011}{2010}\)

b) Ta có: 100 số hạng của dãy đều bé hơn 1/100

\(\Rightarrow\)\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}< \frac{1}{100}\cdot100\)

Hay \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}< 1\)

1/101 + 1/102 + .........+ 2009/2010 + 1/200 > 5/8

vì 2009/2010>5/8

a, Xét 2010 . 2010 = (2009+1).2010 

= 2009.2010 +2010

= (2009.2010+2009)+1

= 2009.(2010+1)+1

= 2009.2011+1 

>= 2009.2010

=> 2010/2009 > 2011/2010

Tk mk nha

a, \(\frac{2010}{2009}\)và \(\frac{2011}{2010}\)

Ta có:

2010.2010 = ( 2009 + 1 ) . 2010

                  = 2009 . 2010 + 2010

                  = ( 2009 . 2010 + 2019 ) + 1

                  = 2019 . ( 2010 + 1 ) + 1

                  = 2019 . 2011 + 1

\(\Rightarrow\)\(\frac{2010}{2009}>\frac{2011}{2010}\)

b, \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...........+\frac{1}{200}\)và 1

Ta có:

\(\frac{1}{101}< 1;\frac{1}{102}< 1;\frac{1}{103}< 1;........;\frac{1}{200}< 1\)

\(\Rightarrow\)\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+.............+\frac{1}{200}< 1\)

c) P = \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)

\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}\right)\)

Dễ thấy \(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}>\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\)(50 hạng tử)

\(\Leftrightarrow\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}>\dfrac{1}{150}.50=\dfrac{1}{3}\)(1)

Tương tự

 \(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}\)(50 hạng tử)

\(\Leftrightarrow\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}>50.\dfrac{1}{200}=\dfrac{1}{4}\)(2) 

Từ (1) và (2) ta được

\(P>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\) 

P = \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)

\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+...+\dfrac{1}{200}\right)\)

         \(\overline{50\text{ hạng tử }}\)                            \(\overline{50\text{ hạng tử }}\)

\(< \left(\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}\right)+\left(\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\right)\) 

\(=\dfrac{1}{100}.50+\dfrac{1}{150}.50=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)

\(\Rightarrow P< \dfrac{5}{6}< 1\)