Question

[Hoàng Giang]

tính

a, \(\left[\dfrac{17}{2}-\left(\dfrac{-3}{7}+\dfrac{5}{3}\right)\right]\)

b, \(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2012}}{\dfrac{2011}{1}+\dfrac{2010}{2}+\dfrac{2009}{3}+....+\dfrac{1}{2011}}\)

c, \(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.9}+....+\dfrac{1}{44.49}\right)\dfrac{1-3-5-7-...-49}{89}\)

d, 1+\(\dfrac{1}{2}\)(1+2)+\(\dfrac{1}{3}\left(1+2+3\right)+....+\dfrac{1}{20}\left(1+2+....+20\right)\)

e, \(\left[\left(-3,2\right):\dfrac{16}{5}\right]+\left(\dfrac{1}{49}-\dfrac{1}{3^2}\right).\left(\dfrac{1}{49}-\dfrac{1}{4^2}\right).....\left(\dfrac{1}{49}-\dfrac{1}{2022^2}\right)\)

f, 50+\(\dfrac{50}{3}+\dfrac{25}{3}+\dfrac{20}{4}+\dfrac{10}{3}+\dfrac{100}{6.7}+....+\dfrac{100}{98.99}+\dfrac{1}{99}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(\left[\dfrac{17}{2}-\left(-\dfrac{3}{7}+\dfrac{5}{3}\right)\right]=\dfrac{17}{2}+\dfrac{3}{7}-\dfrac{5}{3}\)

\(=\dfrac{17\cdot21+3\cdot6-5\cdot14}{42}=\dfrac{305}{42}\)

b: \(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2012}}{\dfrac{2011}{1}+\dfrac{2010}{2}+...+\dfrac{1}{2011}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}}{\left(1+\dfrac{2010}{2}\right)+\left(1+\dfrac{2009}{3}\right)+...+\left(\dfrac{1}{2011}+1\right)+1}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}}{\dfrac{2012}{2}+\dfrac{2012}{3}+...+\dfrac{2012}{2012}}=\dfrac{1}{2012}\)

c: 

\(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{44\cdot49}\)

\(=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{44\cdot49}\right)\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\)

\(=\dfrac{1}{5}\cdot\dfrac{45}{196}=\dfrac{9}{196}\)

1-3-5-...-49

Số số hạng trong khoảng từ 3 đến 49 là:

(49-3):2+1=46:2+1=24(số)

Tổng của các số lẻ trong khoảng từ 3 đến 49 là:

\(\left(49+3\right)\cdot\dfrac{24}{2}=12\cdot52=624\)

=>1-3-5-...-49=1-624=-623

\(\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)

\(=\dfrac{9}{196}\cdot\dfrac{-623}{89}=\dfrac{-63}{196}=-\dfrac{9}{28}\)

d: \(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{20}\left(1+2+...+20\right)\)

\(=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{20}\cdot\dfrac{20\cdot21}{2}\)

\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{21}{2}=\dfrac{2+3+4+...+21}{2}\)

\(=\dfrac{\left(21+2\right)\cdot\dfrac{20}{2}}{2}=23\cdot\dfrac{20}{4}=23\cdot5=115\)

e: \(\left[\left(-3,2\right):\dfrac{16}{5}\right]+\left(\dfrac{1}{49}-\dfrac{1}{3^2}\right)\cdot\left(\dfrac{1}{49}-\dfrac{1}{4^2}\right)\cdot...\cdot\left(\dfrac{1}{49}-\dfrac{1}{2022^2}\right)\)

\(=\left(-3,2\right):3,2+\left(\dfrac{1}{49}-\dfrac{1}{49}\right)\cdot\left(\dfrac{1}{49}-\dfrac{1}{9}\right)\cdot...\cdot\left(\dfrac{1}{49}-\dfrac{1}{2022^2}\right)\)

=-1+0

=-1

f: \(50+\dfrac{50}{3}+\dfrac{25}{3}+...+\dfrac{100}{98\cdot99}+\dfrac{1}{99}\)

\(=\dfrac{100}{2}+\dfrac{100}{6}+\dfrac{100}{12}+...+\dfrac{100}{98\cdot99}+\dfrac{100}{99\cdot100}\)

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