\(B=\frac{101}{102}+\frac{102}{103}+\frac{103}{101}\)

\(B=1\)

B < 3

A=\(\frac{1}{100}\)+\(\frac{1}{101}\)+\(\frac{1}{102}\)+...+\(\frac{1}{200}\)

   (Sử dung phương pháp chặn số đầu)

\(\frac{1}{100}\)>\(\frac{1}{101}\)

\(\frac{1}{100}\)>\(\frac{1}{102}\)

           ...

\(\frac{1}{100}\)>\(\frac{1}{200}\)

nên \(\frac{1}{100}\)+\(\frac{1}{101}\)+\(\frac{1}{102}\)+...+\(\frac{1}{200}\)> \(\frac{1}{100}\)+\(\frac{1}{100}\)+...+\(\frac{1}{100}\)(có 101 phân số)

\(\Rightarrow\)\(\frac{1}{100}\)+\(\frac{1}{101}\)+\(\frac{1}{102}\)+...+\(\frac{1}{200}\)>101.\(\frac{1}{100}\)=\(\frac{101}{100}\)>1>\(\frac{3}{4}\)

\(\Rightarrow\)A >\(\frac{3}{4}\)

\(101A=\frac{101\left(101^{102}+1\right)}{101^{103}+1}=\frac{101^{103}+101}{101^{103}+1}=\frac{101^{103}+1+100}{101^{103}+1}=\frac{101^{103}+1}{101^{103}+1}+\frac{100}{101^{103}+1}=1+\frac{100}{100^{103}+1}\)

\(101B=\frac{101\left(101^{103}+1\right)}{101^{104}+1}=\frac{101^{104}+101}{101^{104}+1}=\frac{101^{104}+1+100}{101^{104}+1}=\frac{101^{104}+1}{101^{104}+1}+\frac{100}{101^{104}+1}=1+\frac{100}{101^{104}+1}\)

vì 100¹⁰³+1<100¹⁰⁴+1

=>\(\frac{100}{100^{103}+1}>\frac{100}{100^{104}+1}\)

=>\(1+\frac{100}{100^{103}+1}>1+\frac{100}{100^{104}+1}\)

=>A>B

Giải:

Ta gọi:

A=101/102+102/103

B=101+102/102+103

Ta có:

B=101+102/102+103

B=101/102+103+102/102+103

Vì 101/102+103<101/102

    102/102+103<102/103

nên A>B

Chúc bạn học tốt!

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

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

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

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

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

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

b) Ta có: \(\frac{1}{101}>0\)

              \(\frac{1}{102}>0\)

                ...............,....

                 \(\frac{1}{200}>0\)

\(\Rightarrow S>0\left(1\right)\)

Lại có: \(\frac{1}{101}< \frac{1}{100}\)

             \(\frac{1}{102}< \frac{1}{100}\)

               ......................

             \(\frac{1}{200}< \frac{1}{100}\)

\(\Rightarrow S< \frac{1}{100}.100\)

\(\Rightarrow S< 1\left(2\right)\)

Từ (1) và (2) \(\Rightarrow0< S< 1\)

Vậy S ko là   số tự nhiên

a, ta có 1/101<1/100; 1/102<1/100;...;1/109<1/100

=> S=1/101+1/102+...+1/109< 1/100+1/100+...+1/100=9/100

=>S<9/100

b,ta thấy S luôn >0

S=1/101+1/102+...+1/200<1/100+1/100+...+1/100=1

=>S<1

=>0<S<1 => S không phải số tự nhiên

\(\frac{1}{101}< \frac{1}{100};\frac{1}{102}< \frac{1}{100};\frac{1}{103}< \frac{1}{100};......;\frac{1}{109}< \frac{1}{100}\)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+....+\frac{1}{109}< \frac{1}{100}+\frac{1}{100}+\frac{1}{100}+....+\frac{1}{100}\)

\(\Rightarrow S< 9\cdot\frac{1}{100}\)

\(\Rightarrow S< \frac{9}{100}\)

Vậy \(S< \frac{9}{100}\)

Có:\(\frac{1}{101}< \frac{1}{100}\)

     \(\frac{1}{102}< \frac{1}{100}\)

      ........................

      \(\frac{1}{109}< \frac{1}{100}\)

=>\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{109}< \frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)

                                                                               (9 phân số)

\(=>\frac{1}{101}+\frac{1}{102}+...+\frac{1}{109}< \frac{9}{100}\)

Vậy thì sửa lại đề là \(\frac{102}{103}\) và \(\frac{103}{104}\)

Bg

Ta có: \(\text{​​}\frac{102}{103}+\frac{1}{103}=1\)và \(\frac{103}{104}+\frac{1}{104}=1\)

Vì \(\frac{1}{103}>\frac{1}{104}\)

Nên \(\frac{102}{103}< \frac{103}{104}\)

Vậy \(\frac{102}{103}< \frac{103}{104}\)

102/103 + 1/103 = 1  => 102/103 + 2/206 = 1

103/105 +2/105 = 1

2/105 > 2/206

=> 102/103 < 103/105

mình làm đúng rồi nhưng viết nhầm kết luận 

Sửa lại:

Mà 2/206 < 2/105

=>102/103 > 103/105