c: C=1*2+2*3+3*4+...+58*59+59*60

=>3*C=1*2*3+2*3*(4-1)+3*4*(5-2)+...+58*59(60-57)+59*60(61-58)

=>3*C=1*2*3+2*3*4-1*2*3+...+58*59*60-58*59*57+59*60*61-58*59*60

=>3*C=59*60*61

=>C=59*20*61=71980

\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{59.60}\)

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

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

=\(\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{60}\right)-\left(1+\frac{1}{2}+...+\frac{1}{30}\right)=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\)

\(1+\frac{7}{1\cdot2}+\frac{7}{2\cdot3}+\frac{7}{3\cdot4}+...+\frac{7}{59\cdot60}\)

\(=1+7\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{59\cdot60}\right)\)

\(=1+7\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{59}-\frac{1}{60}\right)\)

\(=1+7\left(1-\frac{1}{60}\right)\)

\(=1+7\cdot\frac{59}{60}\)

Cảm ơn bạn nha Sooya.

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

\(A=B\)

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

mik cần giải mà 

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

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

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

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

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

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

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

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

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

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

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

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

\(\text{Mik chỉ pít làm z!!!☺}\)

\(P=\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{59.60}\right).31.32.33....59.60\)

\(\text{Ta có:}\)

\(91=13.7\)

\(\rightarrow4.13+5.17=42.35⋮91\)

\(\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{59.60}\right).31.32.33....59.60\)

\(\rightarrow\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{59.60}\right).31.32.....60.42.35\)

\(\rightarrow\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{59.60}\right).31.32....60.20.91⋮91\)

Đặt: \(\left\{{}\begin{matrix}A=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{60}\\B=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{59.60}\end{matrix}\right.\)

Ta có:

\(B=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{59.60}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{59}-\dfrac{1}{60}\)

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

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{60}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{60}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{60}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{30}\right)\)

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

\(\Rightarrow B=A\)

Vậy \(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{59.60}\) (Đpcm)

Ta có:

\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+......+\dfrac{1}{59.60}\)

= \(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+......+\dfrac{1}{59}-\dfrac{1}{60}\)

= \(\left(1+\dfrac{1}{3}+\dfrac{1}{5}+....+\dfrac{1}{59}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{60}\right)\)

- \(2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+....+\dfrac{1}{60}\right)\)

= \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{60}\right)\) - \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{30}\right)\)

=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{30}\right)\)+ \(\left(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{60}\right)\)

- \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{30}\right)\)

= \(\left(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{60}\right)\)

Vậy\(\left(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{60}\right)\)= \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+....+\dfrac{1}{59.60}\)

\(B=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{59.60}\)

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

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

\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{59}+\frac{1}{60}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)

\(B=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{30}\right)\)

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

A > B nhé tích mk vs

cậu giải giúp tớ cái