\(30A=\frac{30^{32}+30}{30^{32}+1}=\frac{30^{32}+1+29}{30^{32}+1}=1+\frac{29}{30^{32}+1}\)

\(30B=\frac{30^{33}+30}{30^{33}+1}=\frac{30^{33}+1+29}{30^{33}+1}=1+\frac{29}{30^{33}+1}\)

Vì \(\frac{29}{30^{32}+1}>\frac{29}{30^{33}+1}\) nên \(1+\frac{29}{30^{32}+1}>1+\frac{29}{30^{33}+1}\Rightarrow30A>30B\Rightarrow A>B\)

Vậy \(A>B.\)

Chúc bạn học tốt.

s<2/3

\(\text{Có 3 trường hợp có thể xảy ra:}\)

\(A=B\)

\(A< B\)
\(A>B\)

mik cần giải mà 

\(A=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)

\(Mà:\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}>\frac{1}{40}.10=\frac{1}{4}\left(\text{10 số hạng}\right)\)

\(\text{Tương tự}:\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}>\frac{1}{5}\)

\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}>\frac{1}{6}\)

\(\Rightarrow A>\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)

\(\Rightarrow A>\frac{37}{60}\)

\(Mà\)\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}>\frac{3}{5}\)

\(\Rightarrow A>\frac{3}{5}\)

\(A=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)

\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}< \frac{1}{31}.10=\frac{10}{30}=\frac{1}{3}\left(\text{10 số hạng}\right)\)

\(\Rightarrow A< \frac{4}{5}\)

\(\Rightarrow\frac{3}{5}< A< \frac{4}{5}\)

\(\text{Mik chỉ pít làm z!!!☺}\)

ớ chết, mk nhầm, lm lại nha

\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)

\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)

\(S< \frac{1}{30}.10+\frac{1}{40}.10+\frac{1}{50}.10\)

\(S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}< \frac{4}{5}\)

=> \(S< \frac{4}{5}\)

\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)

\(S< 30.\frac{1}{60}\)

\(S< \frac{1}{2}< \frac{4}{5}\)

\(S< \frac{4}{5}\)

\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)

\(=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)

\(\Rightarrow S< \frac{1}{30}.10+\frac{1}{40}.10+\frac{1}{50}.10\)

\(S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}< \frac{4}{5}\)

\(V\text{ậy}:S< \frac{4}{5}\)

A>4/5

B = 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/59.60

B = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... + 1/59 - 1/60

B = (1 + 1/3 + 1/5 + ... + 1/59) - (1/2 + 1/4 + 1/6 + ... + 1/60)

B = (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/59 + 1/60) - 2.(1/2 + 1/4 + 1/6 + ... + 1/60)

B = (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... + 1/59 + 1/60) - (1 + 1/2 + 1/3 + ... + 1/30)

B = 1/31 + 1/32 + 1/33 + ... + 1/60 = A

=> B = A

ta có: Lớn nhất của A là:\(\frac{1}{31}+\frac{1}{31}+...+\frac{1}{31}\)(30 phân số)

         =30/31

  B=1-\(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{3}+...+\frac{1}{59}-\frac{1}{60}\)\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{59}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{60}\right)\)

Bé nhất của của B là :\(\left(1+1+...+1\right)-\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)\)

                                \(=30-\frac{30}{60}\)

=>B>A