mà mình không phải là Thiên Tỷ mình tên là Gia Yến đấy nhé

a) \(\dfrac{2}{1\times4}+\dfrac{2}{4\times7}+\dfrac{2}{7\times10}+...+\dfrac{2}{97\times100}\)

\(=2.\left(\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+\dfrac{1}{7\times10}+...+\dfrac{1}{97\times100}\right)\)

\(=2.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(=2.\left(1-\dfrac{1}{100}\right)\)

\(=2.\dfrac{99}{100}\)

\(=\dfrac{99}{50}\)

_____

b) \(\dfrac{3}{1\times5}+\dfrac{3}{5\times9}+\dfrac{3}{9\times13}+...+\dfrac{3}{97\times101}\)

\(=3.\left(\dfrac{1}{1\times5}+\dfrac{1}{5\times9}+\dfrac{1}{9\times13}+...+\dfrac{1}{97\times101}\right)\)

\(=3.\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{97}-\dfrac{1}{101}\right)\)

\(=3.\left(1-\dfrac{1}{101}\right)\)

\(=3.\dfrac{100}{101}\)

\(=\dfrac{300}{101}\)

dấu " . "  là dấu nhân nha bn, nãy viết nhanh quên lớp 

\(\frac{11}{1.4}+\frac{11}{4.7}+...+\frac{11}{100.103}\)

\(=\frac{11}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

\(=\frac{11}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\right)\)

\(=\frac{11}{3}\left(1-\frac{1}{103}\right)\)

Tự tính

\(\frac{11}{1.4}+\frac{11}{4.7}+...+\frac{11}{100.103}\)

= \(\frac{11}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\right)\)

= \(\frac{11}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)\)

= \(\frac{11}{3}.\left(1-\frac{1}{103}\right)\)

= \(\frac{11}{3}.\frac{102}{103}\)

= \(\frac{374}{103}\)

Đặt \(A=\dfrac{2}{1.5}+\dfrac{3}{5.11}+\dfrac{4}{11.19}+\dfrac{5}{19.29}+\dfrac{6}{29.41}\)

\(2A=\dfrac{4}{1.5}+\dfrac{6}{5.11}+\dfrac{8}{11.19}+\dfrac{10}{19.29}+\dfrac{12}{29.41}\)

\(2A=\dfrac{1}{1}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{29}+\dfrac{1}{29}-\dfrac{1}{41}\)

\(2A=\dfrac{1}{1}-\dfrac{1}{41}=\dfrac{40}{41}\)

\(A=\dfrac{40}{41}:2=\dfrac{20}{41}\)

Đặt \(B=\dfrac{1}{1.4}+\dfrac{2}{4.10}+\dfrac{3}{10.19}+\dfrac{4}{19.31}\)

Tương tự ta dễ dàng tình đc:

\(B=\dfrac{10}{31}=\dfrac{20}{62}\)

Ta có: \(\dfrac{20}{41}< \dfrac{20}{62}\Rightarrow A< B\)

Vậy ........................

Chúc bn học tốt

Đặt A=2/1.5+3/5.11+...+6/29.41

B=1/1.4+2/4.10+3/10.19+4/19.31

2A=4/1.5+6/5.11/+...+12/29.41

2A=1-1/5+1/5-1/11+...+1/29-1/41

2A=1-1/41

2A=40/41

A=20/41

3B=3/1.4+6/4.10+9/10.19+12/19.31

3B=1-1/4+1/4-1/10+1/10-1/19+1/19-1/31

3B=1-1/31

3B=30/31

B=10/31

Vì 10/31<20/41 nên B<A

đáp án là 277.1937

277.1937

giúp mình với ạ

Bài 1:

$M=3.4.5+4.5.6+...+13.14.15$

$4M=3.4.5(6-2)+4.5.6(7-3)+....+13.14.15(16-12)$

$=-2.3.4.5+3.4.5.6-3.4.5.6+4.5.6.7+....-12.13.14.15+13.14.15.16$

$=-2.3.4.5+13.14.15.16=43560$

$M=43560:4=10890$

Bài 2:

a.

$3M=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}$

$=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{100-97}{97.100}$

$=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}$

$=1-\frac{1}{100}=\frac{99}{100}$

$M=\frac{99}{100}:3=\frac{33}{100}$