mình không biết làm bài 2

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

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

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

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

\(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{59\cdot60}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{69}-\frac{1}{60}\)

\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{59}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{25}\)

\(A=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)

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

\(A=B\)

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

mik cần giải mà 

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

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

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

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

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

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

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

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

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

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

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

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

\(\text{Mik chỉ pít làm z!!!☺}\)

\(A=1.2+2.3+3.4+4.5+...+59.60\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+...+59.60.\left(61-58\right)\)

\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+...+59.60.61-58.59.60\)

\(\Rightarrow3A=59.60.61\)

\(\Rightarrow A=\frac{59.60.61}{3}\)

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{59.60}\)

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

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

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

\(=1-\dfrac{1}{60}=\dfrac{59}{60}\)