cho A=1/3+1/6+1/10+1/15+........+1/91+1/105
So sánh A với 1
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A= [(1+101)x101:2]-(102-103)
A= 5151+1
A=5152
B= [1+(-3)]+[4+(-5)]+.......[101+(-103)]+105
B= (-2)+(-2)...........+(-2)+105
=> A>B
B=(-2)x26+105
B=(-56)+105
B= 49
cái => A>B nó nằm ở dưới cùng ấy. Nãy gõ chứ nó bị nhảy phím
A=2(1/2+1/6+...+1/90)
=2(1-1/2+1/2-1/3+...+1/9-1/10)
=2*9/10=9/5<2
A = \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + ..+ \(\dfrac{1}{55}\)+ \(\dfrac{1}{66}\)
A = 2 \(\times\) ( \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\) + \(\dfrac{1}{30}\) +...+ \(\dfrac{1}{110}\) + \(\dfrac{1}{132}\))
A = 2 \(\times\) ( \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\)+ \(\dfrac{1}{5.6}\) +...+ \(\dfrac{1}{10.11}\)+ \(\dfrac{1}{11.12}\))
A = 2 \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) +...+ \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)+ \(\dfrac{1}{11}\) - \(\dfrac{1}{12}\))
A = 2 \(\times\) ( \(\dfrac{1}{2}\) - \(\dfrac{1}{12}\))
A = 1 - \(\dfrac{1}{6}\) < 1
Vậy A = \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + \(\dfrac{1}{15}\) + ...+ \(\dfrac{1}{55}\)+ \(\dfrac{1}{66}\) < 1
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}>1\)
\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{66}\)
\(\frac{A}{2}=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\)
\(\frac{A}{2}=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{11\cdot12}\)
\(\frac{A}{2}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{11}-\frac{1}{12}\)
\(\frac{A}{2}=\frac{1}{4}-\frac{1}{12}\)
\(\Rightarrow A=\frac{2}{4}-\frac{2}{12}=\frac{16}{48}\)
\(B=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{55}\)
\(\frac{B}{2}=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}\)
\(\frac{B}{2}=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{10\cdot11}\)
\(\frac{B}{2}=\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)
\(\frac{B}{2}=\frac{1}{3}-\frac{1}{11}\)
\(\Rightarrow B=\frac{2}{3}-\frac{2}{11}=\frac{16}{33}\)
Mà \(\frac{16}{48}< \frac{16}{33}\Rightarrow A< B\)
Vậy : A < B
A=\(\frac{10^8+2}{10^8-1}=1+\frac{3}{10^8-1}\)
\(B=\frac{10^8}{10^8-3}=1+\frac{3}{10^8-3}\)
Vì\(10^8-1>10^8-3\)
\(\Rightarrow\frac{3}{10^8-1}< \frac{3}{10^8-3}\)
\(\Rightarrow1+\frac{3}{10^8-1}< 1+\frac{3}{10^8-3}\)
Vậy \(A< B\)
nhân 2 vế với 1/2 ta có
1/2 x A = 1/2 x (1/3 + 1/6 +1/10 + 1/15 + .......+1/91 + 1/105 )
1/2 x A = 1/6 +1/12 + 1/20 + 1/30 + ...............+1/182 + 1/210
1/2 x A = 1/(2x3) + 1/(3x4) + 1/(4x5) + 1/(5x6) +................+1/(13x14) + 1/(14x15)
1/2 x A = 1/2 - 1/3 +1/3 -1/4 + 1/4 - 1/5 +1/5 - 1/6+.........+1/13 - 1/14 + 1/14 - 1/15
1/2 x A = 1/2 - 1/15 =13/30
=> A = 13/30 : 1/2=13/15 <1