\(\frac{1}{M}=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+...+\frac{1}{\frac{59.60}{2}}\)

\(\frac{1}{M}=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{59.60}\)

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

\(\frac{1}{M}=\frac{2}{3}-\frac{2}{60}< \frac{2}{3}\)

-theo t đề là M chứ ko phải 1/M 

Ta có : \(\frac{1}{m}=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{59.60}=2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{60}\right)=\frac{19}{30}\)

\(\Rightarrow m=\frac{30}{19}>\frac{2}{3}\)

\(Tac\text{ó}:\frac{1}{m}=\frac{2}{3.4}+\frac{2}{4.5}+.....+\frac{2}{59.60}=2\left(\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{59}-\frac{1}{60}\right)\)

\(=>2\left(\frac{1}{3}-\frac{1}{60}\right)=\frac{19}{30}\\ =>m=\frac{30}{19}>\frac{2}{3}\)

\(\frac{1}{M}=\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}+...+\frac{1}{1+2+3+...+59}\)

\(\frac{1}{M}=\frac{1}{3\left(1+3\right):2}+\frac{1}{4\left(1+4\right):2}+\frac{1}{5\left(1+5\right):2}+...+\frac{1}{59\left(1+59\right):2}\)

\(\frac{1}{M}=\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{59.60}\)

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

\(\frac{1}{M}=2\left(\frac{1}{3}-\frac{1}{60}\right)\)

\(\frac{1}{M}=\frac{1}{2}.\frac{19}{60}\)

\(\frac{1}{M}=\frac{19}{120}\)

\(M=\frac{120}{19}>\frac{2}{3}\left(đpcm\right)\)

chuẩn men

\(M=\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+59}\)

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

\(M=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+\frac{1}{\frac{5.6}{2}}+...+\frac{1}{\frac{59.60}{2}}\)

\(M=\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{59.60}\)

\(M=2.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{59.60}\right)\)

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

\(M=2.\left(\frac{1}{3}-\frac{1}{60}\right)\)

\(M< 2.\frac{1}{3}\)

\(M< \frac{2}{3}\)

Mik lười quá bạn tham khảo câu 3 tại đây nhé:

Câu hỏi của nguyen linh nhi - Toán lớp 6 - Học toán với OnlineMath

\(S=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\)

\(2S=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}-\frac{1}{38\cdot39}\)

\(2S=\frac{1}{2}-\frac{1}{38\cdot39}\)

\(S=\frac{1}{4}-\frac{1}{2\cdot38\cdot39}< \frac{1}{4}\)

Ta có: 

\(1+2+3=\frac{\left(1+3\right).3}{2}=\frac{4.3}{2}\)

\(1+2+3+4=\frac{\left(1+4\right).4}{2}=\frac{5.4}{2}\)

\(.................\)

\(1+2+3+...+59=\frac{\left(1+59\right).59}{2}=\frac{60.59}{2}\)

Nên : \(M=\frac{1}{\frac{4.3}{2}}+\frac{1}{\frac{5.4}{2}}+...+\frac{1}{\frac{60.59}{2}}\) , suy ra :\(2M=\frac{1}{4.3}+\frac{1}{5.4}+...+\frac{1}{60.59}\)

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

                                                                               \(2M=\frac{1}{3}-\frac{1}{60}<\frac{1}{3}\)

Do đó :                                                                        \(M<\frac{1}{3}.2\)

                                                                                   \(M<\frac{2}{3}\)

Mình sửa lại \(2M\) thành 1/2 M nhé

Xét Tổng \(A=1+2+3+4+......+n\)Ta có: \(A=\frac{n\left(n+1\right)}{2}\)Vậy \(\frac{1}{A}=\frac{2}{n\left(n+1\right)}=2\left(\frac{1}{n}-\frac{1}{n+1}\right)\)

Vậy \(\frac{1}{1+2+3}=\frac{1}{6}=2\left(\frac{1}{3}-\frac{1}{4}\right)\)

\(\frac{1}{1+2+3+4}=\frac{1}{10}=2\left(\frac{1}{4}-\frac{1}{5}\right)\)

Tương tự:....................................
\(\frac{1}{1+2+3+4+.......+59}=\frac{1}{1770}=2\left(\frac{1}{59}-\frac{1}{60}\right)\)

Vậy \(M=2\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.................+\frac{1}{59}-\frac{1}{60}\right)\)Hay \(M=2\left(\frac{1}{3}-\frac{1}{60}\right)=\frac{19}{30}<\frac{2}{3}\)