So sánh
A=\(\frac{30^{31}+1}{30^{32}+1}\) và B=\(\frac{30^{32}+1}{30^{33}+1}\)
K
Khách
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VK
1
28 tháng 3 2017
A)
\(\frac{1}{30}\)-\(\frac{1}{31}\)+\(\frac{1}{31}\)-\(\frac{1}{32}\)+\(\frac{1}{32}\)-\(\frac{1}{33}\)+...+\(\frac{1}{42}\)-\(\frac{1}{43}\)
=\(\frac{1}{30}\)-\(\frac{1}{43}\)
=\(\frac{13}{1290}\)
B)
=\(\frac{2}{2}\)X(\(\frac{1}{3.5}\)+\(\frac{1}{5.7}\)+\(\frac{1}{7.9}\)+\(\frac{1}{9.11}\))
=\(\frac{1}{2}\)X(\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+\(\frac{2}{7.9}\)+\(\frac{2}{9.11}\))
=\(\frac{1}{2}\)X(\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{11}\))
=\(\frac{1}{2}\)X(\(\frac{1}{3}\)-\(\frac{1}{11}\))
=\(\frac{1}{2}\)X\(\frac{8}{33}\)
=\(\frac{8}{66}\)=\(\frac{4}{33}\)
NK
2
14 tháng 3 2018
Vì N<1
=> N= 20^31+2/20^32+2
<20^31+2+38/ 20^32+2+38
=20^31+40/ 20^32+40
=20.(20^30+2) / 20.(20^31+2)
=20^30+2 / 20^32+2 = M
Vậy N<M
ND
1
T
2 tháng 12 2016
\(\left(\frac{1}{3}\right)^{30}.x+\left(\frac{1}{3}\right)^{31}=\left(\frac{1}{3}\right)^{32}\)
\(\left(\frac{1}{3}\right)^{30}.\left(x+\frac{1}{3}\right)=\left(\frac{1}{3}\right)^{32}\)
\(x+\frac{1}{3}=\left(\frac{1}{3}\right)^{32}:\left(\frac{1}{3}\right)^{30}\)
\(x+\frac{1}{3}=\left(\frac{1}{3}\right)^2\)
\(x+\frac{1}{3}=\frac{1}{9}\)
\(x=\frac{1}{9}-\frac{1}{3}=\frac{1}{9}-\frac{3}{9}\)
\(x=-\frac{2}{9}\)
R
18 tháng 6 2019
Trả lời
b)(1/3+12/67+13/41)-(79/67-28/41)
=1/3+12/67+13/41-79/67+28/41
=1/3+(12/67-79/67)+(13/41+28/41)
=1/3+(-67/67)+41/41
=1/3+(-1)+1
=1/3+0
=1/3.
13 tháng 5 2019
\(M=\frac{10^{30}+1}{10^{31}+1}\)
\(\Rightarrow10M=\frac{10\cdot(10^{30}+1)}{10^{31}+1}\)
\(\Rightarrow10M=\frac{10^{31}+10}{10^{31}+1}\)
\(\Rightarrow10M=\frac{10^{31}+1+9}{10^{31}+1}\)
\(\Rightarrow10M=1+\frac{9}{10^{31}+1}\)
\(N=\frac{10^{31}+1}{10^{32}+1}\)
\(\Rightarrow10N=\frac{10\cdot(10^{31}+1)}{10^{32}+1}\)
\(\Rightarrow10N=\frac{10^{32}+10}{10^{32}+1}\)
\(\Rightarrow10N=\frac{10^{32}+1+9}{10^{32}+1}\)
\(\Rightarrow10N=1+\frac{9}{10^{32}+1}\)
Mà\(1+\frac{9}{10^{31}+1}>1+\frac{9}{10^{32}+1}\)
Nên \(10M>10N\)
Hay \(M>N\)
\(30A=\frac{30^{32}+30}{30^{32}+1}=\frac{30^{32}+1+29}{30^{32}+1}=1+\frac{29}{30^{32}+1}\)
\(30B=\frac{30^{33}+30}{30^{33}+1}=\frac{30^{33}+1+29}{30^{33}+1}=1+\frac{29}{30^{33}+1}\)
Vì \(\frac{29}{30^{32}+1}>\frac{29}{30^{33}+1}\) nên \(1+\frac{29}{30^{32}+1}>1+\frac{29}{30^{33}+1}\Rightarrow30A>30B\Rightarrow A>B\)
Vậy \(A>B.\)
Chúc bạn học tốt.