\(1.3.5.7.....59=\frac{1.2.3.4.....60}{2.4.6.....60}=\frac{\left(1.2.3.....30\right).\left(31.32.....60\right)}{\left(1.2\right).\left(2.2\right).\left(3.2\right).....\left(30.2\right)}\)

\(=\frac{\left(1.2.3.....30\right).\left(31.32.....60\right)}{\left(1.2.3.....30\right).\left(2.2.....2\right)}=\frac{31}{2}.\frac{32}{2}.....\frac{60}{2}\)

ai trả lời đúng mình tặng coin

S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)

Mà : (1/31+1/32+1/33+...+1/40) > 1/40 x 10 = 1/4 (gồm 10 số hạng)

Tương tự : (1/41 + 1/42 + ...+ 1/50) > 1/5 ;   (1/51 + 1/52+...+1/59+1/60) > 1/6

S > 1/4 + 1/5 + 1/6.

Trong khi đó (1/4 + 1/5 + 1/6) > 3/5

=>S > 3/5                             (1)

S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)

Mà : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)

=> S <  4/5                             (2)

Từ (1) và (2) => 3/5 <S<4/5

         \(\frac{31}{2}\)\(.\)\(\frac{32}{2}\)\(.\)\(\frac{33}{2}\)\(....\)\(\frac{60}{2}\)

\(=\)\(\left[\left(31.32.33....60\right)\right]\)\(.\)\(\left(\frac{1.2.3....30}{2^{30}}\right)\)\(.\)\(\left(1.2.3....30\right)\)

\(=\)\(\left[\frac{\left(1.3.5....59\right).\left(2.4.6....60\right)}{2.4.6....60}\right]\)\(=\)\(1.3.5....59\)

Vậy \(\frac{31}{2}\)\(.\)\(\frac{32}{2}\)\(.\)\(\frac{33}{2}\)\(....\)\(\frac{60}{2}\)\(=\)\(1.3.5....59\)

ta có:Đặt A= \(1.3.5.....59=\frac{1.2.3.4.....59.60}{2.4.6.....60}\)

=\(\frac{1.2.3.....59.60}{2^{30}.\left(1.2.3.....30\right)}=\frac{31.32.....59.60}{2^{30}}\)

= \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\)

vì \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\) = \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\)    

\(\Rightarrow\)A= \(\frac{31}{2}.\frac{32}{2}.....\frac{59}{2}.\frac{60}{2}\)

                                          ( Điều phải chứng minh)

toán nâng cao lớp 6 đấy bạn nha

sửu nhi

:>