So sánh tổng : S = 1/5 + 1/9 + 1/10 + 1/41 + 1/42 với 1/2

S=

=50/50+50/49+50/48+...+50/2

=50.(1/50+1/49+1/48+...+1/4+1/3+1/2)

=50

P=

P=(1/49+1)+(2/48+1)+...+(48/2+1)+1

P= 50/49+50/48+....+50/2+50/50=1

vậy s/p = 1/50

​cho P=1/2+1/3+1/4+...........+1/48+1/49+1/50 và Q=1/49+2/48+3/47+........+47/3+48/2+49/1

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\(S=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\)

\(\Rightarrow S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(\Rightarrow S=1-\dfrac{1}{50}\)

\(\Rightarrow S=\dfrac{49}{50}\)

Phần P bạn xem lại đề

Bài 1:

Ta có:

\(S=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\)

\(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{49}{1}\)

\(\Rightarrow\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{49}{1}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+...+\left(1+\dfrac{48}{2}\right)+1}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}}\)

\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)

Vậy \(\dfrac{S}{P}=\dfrac{1}{50}\)

Bài 2:

Ta có:

\(S=\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}\)

\(=\dfrac{1}{5}+\left(\dfrac{1}{9}+\dfrac{1}{10}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}\right)\)

Nhận xét:

\(\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}\)

\(\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{40}+\dfrac{1}{40}=\dfrac{1}{20}\)

\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)

Vậy \(S=\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{2}\)