Question

Giúp mk

Answer 1

Bài 1:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{119.120}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{120-119}{120}\)

\(=\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+....+\frac{120}{119.120}-\frac{119}{119.120}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{119}-\frac{1}{120}\)

$=1-\frac{1}{120}=\frac{119}{120}$

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$B=\frac{11-10}{10.11}+\frac{12-11}{11.12}+\frac{13-12}{12.13}+....+\frac{1000-999}{999.1000}$

$=\frac{11}{10.11}-\frac{10}{10.11}+\frac{12}{11.12}-\frac{11}{11.12}+\frac{13}{12.13}-\frac{12}{12.13}+...+\frac{1000}{999.1000}-\frac{999}{999.1000}$

$=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{999}-\frac{1}{1000}$

$=\frac{1}{10}-\frac{1}{1000}=\frac{99}{1000}$

Answer 2

Bài 2:

$A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{109-107}{107.109}$

$=\frac{3}{1.3}-\frac{1}{1.3}+\frac{5}{3.5}-\frac{3}{3.5}+\frac{7}{5.7}-\frac{5}{5.7}+...+\frac{109}{107.109}-\frac{107}{107.109}$

$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{107}-\frac{1}{109}$

$=1-\frac{1}{109}=\frac{108}{109}$

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$B=\frac{5-2}{2.5}+\frac{8-5}{5.8}+\frac{11-8}{8.11}+...+\frac{35-32}{32.35}$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{32}-\frac{1}{35}$

$=\frac{1}{2}-\frac{1}{35}=\frac{33}{70}$

Answer 3

Bài 3:

$2A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{398.400}$

$=\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+.....+\frac{400-398}{398.400}$

$=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{398}-\frac{1}{400}$

$=\frac{1}{2}-\frac{1}{400}=\frac{199}{400}$
$A=\frac{199}{400}:2=\frac{199}{800}$

Answer 4

Bài 3:
\(2B=\frac{4}{5.9}+\frac{4}{9.13}+....+\frac{4}{41.45}\)

$=\frac{9-5}{5.9}+\frac{13-9}{9.13}+....+\frac{45-41}{41.45}$

$=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}$

$=\frac{1}{5}-\frac{1}{45}=\frac{8}{45}$

$\Rightarrow B=\frac{8}{45}:2=\frac{4}{45}$

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$\frac{C}{2}=\frac{6}{25.31}+\frac{6}{31.37}+...+\frac{6}{139.145}$

$=\frac{31-25}{25.31}+.\frac{37-31}{31.37}+...+\frac{145-139}{139.145}$

$=\frac{1}{25}-\frac{1}{31}+\frac{1}{31}-\frac{1}{37}+...+\frac{1}{139}-\frac{1}{145}$

$=\frac{1}{25}-\frac{1}{145}=\frac{24}{725}$

$C=2.\frac{24}{725}=\frac{48}{725}$

Answer 5

Bài 4:

$\frac{A}{3}=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+....+\frac{1}{398.400}$

$=\frac{199}{800}$ (kết quả bài 3 phần a)

$\Rightarrow A=3.\frac{199}{800}=\frac{597}{800}$

Answer 6

Bài 5:

$A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{100-99}{99.100}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}$

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

$=\frac{99}{100}$

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$B=\frac{11-10}{10.11}+\frac{12-11}{11.12}+\frac{13-12}{12.13}+...+\frac{200-199}{199.200}$

$=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+....+\frac{1}{199}-\frac{1}{200}$

$=\frac{1}{10}-\frac{1}{200}=\frac{19}{200}$