a.\(\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\)

\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}...\dfrac{100}{99}\)

\(=\dfrac{3.4.5...100}{2.3.4...99}\)

\(=\dfrac{100}{2}=50\)

a,

\(\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)...\left(\dfrac{1}{99}+1\right)\\ =\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{100}{99}\\ =\dfrac{3\cdot4\cdot5\cdot...\cdot100}{2\cdot3\cdot4\cdot...\cdot99}\\ =\dfrac{100}{2}=50\)

b,

\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{100}-1\right)\\ =\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{-3}{4}\cdot...\cdot\dfrac{-99}{100}\\ =\dfrac{\left(-1\right)\left(-2\right)\left(-3\right)\cdot...\cdot\left(-99\right)}{2\cdot3\cdot4\cdot...\cdot100}\\ =\dfrac{\left(-1\right)\left(-1\right)\left(-1\right)\cdot...\left(-1\right)}{100}\left(\text{có }99\text{ số }-1\right)\\ =\dfrac{\left(-1\right)^{99}}{100}\\ =\dfrac{-1}{100}\)

c,

\(C=\dfrac{4}{30}+\dfrac{4}{70}+\dfrac{4}{126}+...+\dfrac{4}{798}\\ =\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{399}\\ =\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{19\cdot21}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\\ =\dfrac{1}{3}-\dfrac{1}{21}\\ =\dfrac{7}{21}-\dfrac{1}{21}\\ =\dfrac{6}{21}=\dfrac{2}{7}\)

b.\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{100}-1\right)\)

\(=\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}...\dfrac{-99}{100}\)

\(=\dfrac{-1.\left(-2\right).\left(-3\right)...\left(-99\right)}{2.3.4...100}\)

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

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

\(A=1-\dfrac{1}{8}=\dfrac{7}{8}\)

\(A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

\(A=\dfrac{367}{420}\approx0,87\)

A = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72

= 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8 + 1/8.9

= 1/1 − 1/2 + 1/2 − 1/3 + 1/3 − 1/4 + 1/4 − 1/5 + 1/5 − 1/6 + 1/6 − 1/7 + 1/7 − 1/8 + 1/8 − 1/9

= 1 − 1/9

= 8/9

a) \(A=\dfrac{3}{5}+6\dfrac{5}{6}+\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)

\(=\dfrac{3}{5}+\dfrac{41}{6}\left(11\dfrac{1}{4}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)

\(=\dfrac{3}{5}+\dfrac{41}{6}.2.\dfrac{3}{25}\)

\(=\dfrac{3}{5}+\dfrac{41}{25}\)

\(=\dfrac{15}{25}+\dfrac{41}{25}\)

\(=\dfrac{56}{25}\)

a) 

b) \(B=\dfrac{-1}{2}+\dfrac{-1}{6}+\dfrac{-1}{12}+\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)

\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\) \(1-\dfrac{1}{10}\)

\(=\dfrac{-9}{10}\)

Đặt `A=(1-3x)/(2x)+(3x-2)/(2x-1)+(3x-2)/(2x-4x^2)`

`=(2x(3x-2))/(2x(2x-1))-((3x-1)(2x-1))/(2x(2x-1))-(3x-2)/(2x(2x-1))`

`=(6x^2-4x-6x^2+5x-1-3x+2)/(2x(2x-1))`

`=(-2x+1)/(2x(2x-1))`

`=-1/(2x)`

`2x=1/(483)`

`=>A=-1/(1/483)=-483`

\(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\)

⇔\(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x\left(1-2x\right)}\)

⇔\(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{(3x-2)}{2x\left(2x-1\right)}\)

ĐKXĐ: \(\left\{{}\begin{matrix}2x\ne0\\2x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\2x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne\dfrac{1}{2}\end{matrix}\right.\)

MTC: 2x(2x-1)

\(\dfrac{1-3x\left(2x-1\right)}{2x\left(2x-1\right)}+\dfrac{(3x-2)(2x)}{(2x-1)(2x)}-\dfrac{(3x-2)}{2x\left(2x-1\right)}\)

\(\Rightarrow1-3x\left(2x-1\right)+\left(3x-2\right)\left(2x\right)-\left(3x-2\right)\)

\(\Leftrightarrow1-6x^2+3x+6x^2-4x-3x+2\\ \Rightarrow-4x+3\)

Thay \(x=\dfrac{1}{966}\)vào biểu thức trên

ta có -4x+3= \(-4\times\dfrac{1}{966}+3=\dfrac{-4}{966}+3=\dfrac{-2}{483}+3=\dfrac{-2}{483}+\dfrac{1449}{483}=\dfrac{1447}{483}\)

= 3/10 + 4/7 . 5/4 - 1/10 
= 3/10 + 5/7 - 1/70 
= 1.

\(\dfrac{3}{10}+\dfrac{4}{7}.\dfrac{5}{4}-\dfrac{1}{70}=\dfrac{3}{10}+\dfrac{5}{7}-\dfrac{1}{70}=\dfrac{21}{70}+\dfrac{50}{70}-\dfrac{10}{70}=\dfrac{70}{70}=1\)

1) \(=\dfrac{1}{2}.\dfrac{3}{5}+\dfrac{4}{7}.\dfrac{5}{4}-\dfrac{1}{70}\)

\(=\dfrac{3}{10}+\dfrac{5}{7}-\dfrac{1}{70}\)

\(=\dfrac{21}{70}+\dfrac{50}{70}-\dfrac{1}{70}\)

\(=1\)

2)

\(=\left(\dfrac{30}{45}+\dfrac{9}{45}-\dfrac{20}{45}\right):\left(\dfrac{15}{45}+\dfrac{18}{45}-\dfrac{30}{45}\right)\)

\(=\dfrac{19}{45}:\dfrac{1}{45}\)

\(=\dfrac{19}{45}.45\)

\(=19\)

11,6

= 1,1 + 1,2 + 1,3 + 1,4 + 1,5 + 1,6 + 1,7 + 1,8 + 1,9
= 2,3 + 1,3 + 1,4 + 1,5 + 1,6 + 1,7 + 1,8 + 1,9
= 3,6 + 1,4 + 1,5 + 1,6 + 1,7 + 1,8 + 1,9
= 5 + 1,5 + 1,6 + 1,7 + 1,8 + 1,9
= 6,5 + 1,6 + 1,7 + 1,8 + 1,9
= 8,1 + 1,7 + 1,8 + 1,9
= 9,8 + 1,8 + 1,9
= 11,6 + 1,9
= 13,5