So sánh A với B
A=\(\frac{13^{2000}+1}{13^{2001}+1}\)
B=\(\frac{13^{2001}+1}{13^{2002}+1}\)
K
Khách
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13 tháng 10 2019
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
nên x + 1 = 0 => x = -1
Vậy x = -1
b) \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(1+\frac{x+4}{2000}+1+\frac{x+3}{2001}=1+\frac{x+2}{2002}+1+\frac{x+1}{2003}\)
\(\frac{2004+x}{2000}+\frac{2004+x}{2001}=\frac{2004+x}{2002}+\frac{2004+x}{2003}\)
\(\frac{2004+x}{2000}+\frac{2004+x}{2001}-\frac{2004+x}{2002}-\frac{2004+x}{2003}=0\)
\(\left(2004+x\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Mà \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\)
nên 2004 + x = 0 => x = -2004
Vậy x = -2004
=))
NT
1
20 tháng 6 2018
13/27 và 7/15
\(\frac{13}{27}\) = 1:\(\frac{27}{13}\)= 1: \(\frac{26+1}{13}\) = 1: ( 2+\(\frac{1}{13}\))
\(\frac{7}{15}\)= 1:\(\frac{15}{7}\)= 1: \(\frac{14+1}{7}\)= 1: ( 2+ \(\frac{1}{7}\))
ta có \(\frac{1}{13}\)< \(\frac{1}{7}\)=> 2+\(\frac{1}{13}\)< 2+ \(\frac{1}{7}\) => 1: ( 2+\(\frac{1}{13}\)) > 1: ( 2+ \(\frac{1}{7}\))
vậy \(\frac{13}{27}\)>\(\frac{7}{15}\)- 2000/2001 và 2001/2002
\(\frac{2000}{2001}\)= \(\frac{2001-1}{2001}\)= 1 - \(\frac{1}{2001}\)
\(\frac{2001}{2002}\)= \(\frac{2002-1}{2002}\)= 1 - \(\frac{1}{2002}\)
ta có \(\frac{1}{2001}\)> \(\frac{1}{2002}\) => 1 - \(\frac{1}{2001}\) < 1 - \(\frac{1}{2002}\)
vậy \(\frac{2000}{2001}\)< \(\frac{2001}{2002}\)
VM
3
ML
23 tháng 5 2016
a.\(\frac{13}{17}\)=1-\(\frac{4}{17}\); \(\frac{46}{50}\)=1-\(\frac{4}{50}\)
Vì \(\frac{4}{17}\)>\(\frac{4}{50}\)=> 1-\(\frac{4}{17}\)<1-\(\frac{4}{50}\)
Vậy\(\frac{13}{17}\)<\(\frac{46}{50}\)
QH
1
19 tháng 4 2015
Ta có:
B = \(\frac{2000}{2001+2002}\)+ \(\frac{2001}{2001+2002}\)
Vì \(\frac{2000}{2001}\)> \(\frac{2000}{2001+2002}\)
\(\frac{2001}{2002}\)> \(\frac{2001}{2001+2002}\)
=> \(\left(\frac{2000}{2001}+\frac{2001}{2002}\right)\)> \(\left(\frac{2000}{2001+2002}+\frac{2001}{2001+2001}\right)\)
=> A>B
Vậy A>B
PB
1
VT
26 tháng 3 2016
\(\frac{x+16}{2000}+1+\frac{x+15}{2001}+1=\frac{x+14}{2002}+1+\frac{x+13}{2003}+1\)
\(\frac{x+2016}{2000}+\frac{x+2016}{2001}=\frac{x+2016}{2002}+\frac{x+2016}{2003}\)
\(\frac{x+2016}{2000}+\frac{x+2016}{2001}-\frac{x+2016}{2002}-\frac{x+2016}{2003}=0\)
\(\left(x+2016\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(x-2016=0\)
\(x=2016\)
Vì \(13^{2001}+1< 13^{2002}+1\) nên \(B=\frac{13^{2001}+1}{13^{2002}+1}< 1\)
\(\Rightarrow B=\frac{13^{2001}+1}{13^{2002}+1}< \frac{13^{2001}+1+12}{13^{2002}+1+12}=\frac{13^{2001}+13}{13^{2002}+13}=\frac{13\left(13^{2000}+1\right)}{13\left(13^{2001}+1\right)}=\frac{13^{2000}+1}{13^{2001}+1}=A\)
\(\Rightarrow B< A\)