Cho A=1/2!+2/3!+3/4!+...+9/10!.So sánh A với 1
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NV
0
NC
28 tháng 8 2020
Bài làm:
Ta có: \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}\)
\(A=\left(1+\frac{1}{3}+...+\frac{1}{9}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
\(A=\left[\left(1+\frac{1}{3}+...+\frac{1}{9}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\right]-\left[\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\right]\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)=B\)
Vậy A = B
16 tháng 6 2017
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A=\left(1+\frac{1}{3}+...+\frac{1}{9}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}\right)-\left(1+\frac{1}{2}+...+\frac{1}{5}\right)\)
\(A=\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...+\frac{1}{10}\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{5}\right)\)
Vậy A = B và A = 1/6 + 1/7 + 1/8 + 1/9 + 1/10
TH
16 tháng 6 2017
1/ A= \(\left(\frac{1}{1.2}\right)+\left(\frac{1}{3.4}\right)+...+\left(\frac{1}{9.10}\right)\)
B=(1/1+1/2+1/3+...+1/10)- (1/1+1/2+...+1/5)
<=> B=1/6+1/7+1/8+1/9+1/10.
PN
10 tháng 5 2021
a,\(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
\(=>5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
\(=>5A-A=1-\frac{1}{5^{100}}=>A=\frac{1-\frac{1}{5^{100}}}{4}\)
b, Ta có \(1-\frac{1}{5^{100}}< 1=>\frac{1-\frac{1}{5^{100}}}{4}< \frac{1}{4}\)hay \(A< \frac{1}{4}\)
19 tháng 3 2017
M = \(\frac{1}{2}\)+ \(\frac{2}{3}\)+ \(\frac{3}{4}\)+\(\frac{4}{5}\)+ \(\frac{5}{6}\)+ \(\frac{6}{7}\)+ \(\frac{7}{8}\)+ \(\frac{8}{9}\)+ \(\frac{9}{10}\)= \(\frac{17819}{2520}\)
Vậy: M > 1
TQ
2
H
15 tháng 7
`3/(-10) ; 1/(-2) ; 4/(-5)=> -3/10 ; -1/2 ; -4/5`
ta có : `-1/2=(-1xx5)/(2xx5)=-5/10 ; -4/5=(-4xx2)/(5xx2)=-8/10`
vậy `3/(-10) < 1/(-2) < 4/(-5)`
`--------------------`
`2/(-10) ; 7/(-5) ; -1/2=>-2/10 ;-7/5;-1/2`
ta có : `-7/5=(-7xx2)/(5xx2)=-14/10; -1/2=(-1xx5)/(2xx5)=-5/10`
vậy `2/(-10) < -1/2 < 7/(-5)`
`---------------------`
`7/(-4) ; -2/5 ; -3/10=> -7/4;-2/5;-3/10`
ta có : `-7/4=(-7xx5)/(4xx5)=-35/20 ; -2/5=(-2xx4)/(5xx4)=-8/20;-3/10=(-3xx2)/(10xx2)=-6/20`
vậy 7/(-4) > -2/5 > -3/10`
Ta có :
\(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{9}{10!}\)
\(A=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{10-1}{10!}\)
\(A=\left(\frac{2}{2!}-\frac{1}{2!}\right)+\left(\frac{3}{3!}-\frac{1}{3!}\right)+\left(\frac{4}{4!}-\frac{1}{4!}\right)+...+\left(\frac{10}{10!}-\frac{1}{10!}\right)\)
\(A=\left(1-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+...+\left(\frac{1}{9!}-\frac{1}{10!}\right)\)
\(A=1-\frac{1}{10!}< 1\)
vậy A < 1 vì \(0< \frac{1}{10!}< 1\)