Ta có:

\(\frac{1}{51}>\frac{1}{100}\)

\(\frac{1}{52}>\frac{1}{100}\)

...

\(\frac{1}{99}>\frac{1}{100}\)

\(\frac{1}{100}=\frac{1}{100}\)

=> S = \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\)

Mà số số hạng của S là: (100 - 51) : 1 + 1 = 50 (số)

=> S \(>\frac{1}{100}.50\)

=> S \(>\frac{1}{2}\)

Vậy S > 1/2.

\(\frac{\frac{4}{17}-\frac{4}{45}+\frac{4}{156}}{\frac{3}{17}-\frac{3}{45}+\frac{3}{156}}=\frac{4.\left(\frac{1}{17}-\frac{1}{45}+\frac{1}{156}\right)}{3.\left(\frac{1}{17}-\frac{1}{45}+\frac{1}{156}\right)}=\frac{4}{3}\)

thanks Le Tai Bao Chau nha

\(\frac{\frac{4}{17}-\frac{4}{45}+\frac{1}{39}}{\frac{3}{17}-\frac{1}{15}+\frac{1}{52}}\)

\(=\frac{\frac{4}{17}-\frac{4}{45}+\frac{4}{156}}{\frac{3}{17}-\frac{3}{45}+\frac{3}{156}}\)

\(=\frac{\frac{1}{17}.4-\frac{1}{45}.4+\frac{1}{156}.4}{\frac{1}{17}.3-\frac{1}{45}.3+\frac{1}{156}.3}\)

\(=\frac{4\left(\frac{1}{17}-\frac{1}{45}+\frac{1}{156}\right)}{3\left(\frac{1}{17}-\frac{1}{45}+\frac{1}{156}\right)}\)

\(=\frac{4}{3}\)

\(45\cdot\left(-\frac{5}{7}\right)+9\cdot5+\frac{8}{9}+135+45\cdot\frac{52}{63}\)
\(=-\frac{225}{7}+45+\frac{8}{9}+135+\frac{260}{7}\)
\(=\left(-\frac{225}{7}+\frac{260}{7}\right)+\left(45+135\right)+\frac{8}{9}\)
\(=\frac{35}{7}+180+\frac{8}{9}\)
\(=5+180+\frac{8}{9}\)
\(=185+\frac{8}{9}=185\frac{8}{9}\)
tick đúng cho tớ nha

a)48*(19+115)+134*52

48 *134 +134*52 

134*(48+52)

134*100

13400

đươc đấy

Ta có:

 A = [15 x (1-1/7-1/12-1/98)] / [ 18 x (1-1/7-1/12-1/98)]

  = 15/18 = 5/6

\(A=\frac{15\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}{18\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}=\frac{15}{18}=\frac{15:3}{18:3}=\frac{5}{6}\)

k cho mk nha

A=15x(1/7-1/12-1/98)/18(1/7-1/12-1/98)

A=15/18

A=5/6

h nhe!!!

\(A=\frac{15\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}{18\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}\)

\(A=\frac{15}{18}=\frac{5}{6}\)

\(A=\frac{15-\frac{15}{7}-\frac{15}{12}-\frac{15}{98}}{18-\frac{18}{7}-\frac{18}{12}-\frac{18}{98}}\)

\(\Rightarrow A=\frac{15\left(\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}{18\left(\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}\)

\(\Rightarrow A=\frac{15}{18}=\frac{5}{6}\)

mk làm chuẩn rồi nha

Giải:

\(S=\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{98}+\dfrac{1}{99}\) 

\(S=\left(\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{74}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{98}+\dfrac{1}{99}\right)\) 

\(\Rightarrow S>\left(\dfrac{1}{50}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{75}+\dfrac{1}{75}\right)\) 

\(\Rightarrow S>\dfrac{1}{2}+\dfrac{1}{3}>\dfrac{1}{2}\) 

\(\Rightarrow S>\dfrac{1}{2}\left(đpcm\right)\) 

Ta có:S=1/50+1/51+1/52+...+1/99

S>1/50+1/50+1/50+....+1/50(50 số hạng)

S>1/50x50

S>1>1/2

=>S>1/2