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\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2021}{2022}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2021}{2022}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2022}\)
=>x+1=2022
hay x=2021
\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}\)
\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{99-98}{98.99}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
\(=1-\dfrac{1}{99}\)
\(A=\dfrac{2021}{2022}=\dfrac{2022-1}{2022}=1-\dfrac{1}{2022}\)
Có \(2022>99>0\Leftrightarrow\dfrac{1}{99}>\dfrac{1}{2022}\)
Suy ra \(A>B\).
đặt A=1/1.2+1/2.3+1/3.4+..........1/49.50
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}<1\)
vậy A<1
1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50
1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50
1 - 1/50 < 1
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}< 1\) (đpcm)
ta có :
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}< 1\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2003\cdot2004}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2003}-\frac{1}{2004}\)
\(=1-\frac{1}{2004}\)
\(=\frac{2004}{2004}+\frac{-1}{2004}=\frac{2003}{2004}\)
#Hoq chắc _ Baccanngon
đặt A=.....
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}\)=\(\frac{2016}{2017}\)
=\(1-\frac{1}{x+1}=\frac{2016}{2017}\)
=\(\frac{x}{x+1}=\frac{2016}{2017}\)
=>x=2016
vậy..............
\(S=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{2021+2022}\)
\(S=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}\)
\(S=\dfrac{1}{2}-\dfrac{1}{2022}\)
\(S=\dfrac{1011}{2022}-\dfrac{1}{2022}\)
\(S=\dfrac{505}{1011}\)
1/1.2+1/2.3+1/3.4+....+1/99.100
=1/1-1/2+1/2-1/3+1/3-1/4+.....+1/99-1/100
=1/1-1/100
=99/100
=>1-1/2+1/2-1/3+...+1/x-1/(x+1)=2022/2021
=>1-1/(x+1)=2022/2021
=>1/(x+1)=-1/2021=1/-2021
=>x+1=-2021
=>x=-2022