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

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

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

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

Ta có: \(B=\frac{101+102+103}{102+103+104}=\frac{101}{102+103+104}+\frac{102}{102+103+104}+\frac{103}{102+103+104}\)

Ta thấy: \(\frac{101}{102}>\frac{101}{102+103+104}\)

              \(\frac{102}{103}>\frac{102}{102+103+104}\)

              \(\frac{103}{104}>\frac{103}{102+103+104}\)

\(\Rightarrow A=\frac{101}{102}+\frac{102}{103}+\frac{103}{104}>\frac{101}{102+103+104}+\frac{102}{102+103+104}+\frac{103}{102+103+104}=\frac{101+102+103}{102+103+104}=B\)

Vậy....

\(B=\frac{101+102+103}{102+103+104}=\frac{101}{102+103+104}+\frac{102}{102+103+104}+\frac{103}{102+103+104}\)

Ta có: \(\frac{101}{102}>\frac{101}{102+103+104}\)

          \(\frac{102}{103}>\frac{102}{102+103+104}\)

          \(\frac{103}{104}>\frac{103}{102+103+104}\)

\(\Rightarrow A=\frac{101}{102}+\frac{102}{103}+\frac{103}{104}>\frac{101}{102+103+104}+\frac{102}{102+103+104}+\frac{103}{102+103+104}=\frac{101+102+103}{102+103+104}=B\)Vậy....

Ta có :\(\frac{101}{102}>\frac{101}{102+103+104}\)

\(\frac{102}{103}>\frac{102}{102+103+104}\)

\(\frac{103}{104}>\frac{103}{102+103+104}\)

Do đó:\(\frac{101}{102}+\frac{102}{103}+\frac{103}{104}>\frac{101+102+103}{102+103+104}\)

Vậy A>B

tk mk nha!!!

\(A=2,970871956;B=\frac{102}{103}\)

\(A>2>1>B\)

\(\Rightarrow A>B\)

Đúng 100%

Đúng 100%

Đúng 100%

Ta có: \(B=\frac{101+102+103}{102+103+104}=\frac{101}{102+103+101}+\frac{102}{102+103+104}+\)\(\frac{103}{102+103+104}\)

Vì: \(\frac{101}{102}>\frac{101}{102+103+104}\)

\(\frac{102}{103}>\frac{102}{102+103+104}\)

\(\frac{103}{104}>\frac{103}{102+103+104}\)

\(\Rightarrow A>B\)

Vậy A > B

A > b nha bạn

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!

Ta thấy mẫu của Ava B bằng nhau vậy chỉ cần so sánh tử mà thôi

mà từ cửa AvaB cũng bằng nhau =>A=B

Tớ thấy mẫu A và B bằng nhau vậy chỉ cần so sánh tử và mẫu.

A và B cũng bằng nhau \(\Rightarrow\) A = B

Học tốt !!!

A và B bằng nhau

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

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

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

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

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

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