Giờ anh đang bận hồi nữa anh giúp cho nha

\(F=\dfrac{1}{x}\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}+\dfrac{1}{100.103}\right)\)

\(3F=\dfrac{1}{x}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}+\dfrac{3}{100.103}\right)\)

\(F=\dfrac{\dfrac{1}{x}\left(\dfrac{1}{3}-\dfrac{1}{103}\right)}{3}=\dfrac{\dfrac{1}{x}.\dfrac{100}{309}}{3}=\dfrac{\dfrac{100x}{309}}{3}=\dfrac{100x}{927}\)

\(F=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{103}\right)=\dfrac{1}{3}.\dfrac{102}{103}=\dfrac{204}{309}\)

\(F=\dfrac{1}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{100\cdot103}\right)\)

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

\(=\dfrac{1}{3}\cdot\dfrac{102}{103}=\dfrac{34}{103}\)

a/ \(\dfrac{3}{11.12}+\dfrac{3}{12.13}+\dfrac{3}{13.14}+\dfrac{3}{14.15}\)

\(=3\left(\dfrac{1}{11.12}+\dfrac{1}{12.13}+\dfrac{1}{13.14}+\dfrac{1}{14.15}\right)\)

\(=3\left(\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\right)\)

\(=3\left(\dfrac{1}{11}-\dfrac{1}{15}\right)\)

\(=\dfrac{4}{55}\)

b/ \(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}\)

\(=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)

\(=\dfrac{2}{3}\)

c/ \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+.....+\dfrac{3}{97.100}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+....+\dfrac{1}{97}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}\)

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

d/ \(\dfrac{3}{2.5}+\dfrac{3}{5.8}+.....+\dfrac{3}{100.103}\)

\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+....+\dfrac{1}{100}-\dfrac{1}{103}\)

\(=\dfrac{1}{2}-\dfrac{1}{103}\)

\(=\dfrac{101}{206}\)

e/ Đặt :

\(A=\dfrac{1}{1.5}+\dfrac{1}{5.10}+....+\dfrac{1}{95.100}\)

\(\Leftrightarrow5A=\dfrac{5}{1.5}+\dfrac{5}{5.10}+....+\dfrac{5}{95.100}\)

\(=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+....+\dfrac{1}{95}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}\)

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

\(\Leftrightarrow A=\dfrac{99}{100}:5=\dfrac{99}{500}\)

Dấu . là dấu nhân nhé <3

\(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-1\right)\\ =\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}-1}\right)\\ =\dfrac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}\\ =\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{1}\\ =\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)

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

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

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

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

Tự tính

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

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

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

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

= \(\frac{11}{3}.\frac{102}{103}\)

= \(\frac{374}{103}\)

\(A=\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{100\cdot103}\)

\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(=\dfrac{98}{515}\)

S=1/1-1/4+1/4+1/7-1/7+1/10+...+1/100-1/103

S=1/1-1/103

S=102/103

Vì 102/103<1 nên S<1

\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{100\cdot103}\)

\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\)

\(S=1-\frac{1}{103}\)

\(S=\frac{102}{103}< 1\)

\(\frac{3}{1x4}+\frac{3}{4x7}+\frac{3}{7x10}+.......+\frac{3}{100x103}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}....+\frac{1}{100}-\frac{1}{103}\)

\(=\frac{1}{1}-\frac{1}{103}\)

=\(\frac{102}{103}\)

Đặt  \(B=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+......+\frac{2}{100\cdot103}\)

\(B=\frac{2}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{100}-\frac{1}{103}\right)\)

\(B=\frac{2}{3}\cdot\left(1-\frac{1}{103}\right)\)

\(B=\frac{2}{3}\cdot\frac{102}{103}\)

\(\Rightarrow B=\frac{68}{103}\)

Đặt \(A=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{100.103}\)

\(A=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(A=\frac{2}{3}\left(1-\frac{1}{103}\right)\)

\(A=\frac{2}{3}\cdot\frac{102}{103}\)

\(A=\frac{68}{103}\)

\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{100.103}\)

\(=\frac{2}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

\(=\frac{2}{3}\cdot\left(1-\frac{1}{103}\right)\)

\(=\frac{2}{3}\cdot\frac{102}{103}=\frac{68}{103}\)

Ta có: \(P=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{11}\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\cdot\dfrac{-10}{11}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{9}{10}\cdot\dfrac{10}{11}\)

\(=\dfrac{1}{11}\)