Ta có :\(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{90}=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{90}\right)\)

                   60 số hạng                                                              30 số hạng                                     30 số hạng

\(>\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)+\left(\frac{1}{90}+\frac{1}{90}+...+\frac{1}{90}\right)=30.\frac{1}{60}+30.\frac{1}{90}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

Vậy \(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{90}>\frac{5}{6}\)

Ta có: \(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=30.\frac{1}{60}=\frac{1}{2}\)

Lại có: \(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}+...+\frac{1}{90}>\frac{1}{90}+\frac{1}{90}+...+\frac{1}{90}=30.\frac{1}{90}=\frac{1}{3}\)

\(\Rightarrow\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{90}>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)

\(\Rightarrow\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{90}>\frac{5}{6}\) (đpcm)

1/31 + 1/32 + 1/33 + ... + 1/90 

= (1/31 + 1/32 + ... + 1/60) + (1/61 + 1/62 + ... + 1/90)

> 1/60 × 30 + 1/90 × 30

> 1/2 + 1/3

> 5/6

(không ghi cách giải)

đáp án : a > 5/6

chúc bn

hok tốt

(ko ghi đề)

đáp án : a > 5 / 6

chúc b

hk tốt

\(B=\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}=\dfrac{30}{60}=\dfrac{1}{2}\)

\(C=\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{90}>\dfrac{1}{90}+\dfrac{1}{90}+...+\dfrac{1}{90}=\dfrac{30}{90}=\dfrac{1}{3}\)

Do đó: \(B+C>\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)(đpcm)