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a) Ta có : 10A = \(\frac{10\left(10^{2004}+1\right)}{10^{2005}+1}=\frac{10^{2005}+10}{10^{2005}+1}=1+\frac{9}{10^{2005}+1}\)
Lại có 10B = \(\frac{10\left(10^{2005}+1\right)}{10^{2006}+1}=\frac{10^{2006}+10}{10^{2006}+1}=1+\frac{9}{10^{2006}+1}\)
Vì \(\frac{9}{10^{2005}+1}>\frac{9}{10^{2006}+1}\Rightarrow1+\frac{9}{10^{2005}+1}>1+\frac{9}{10^{2006}+1}\)
=> 10A > 10B
=> A > B
b) Ta có A = \(\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
Lại có B = \(\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
Vì \(\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\Rightarrow1+\frac{2}{20^{10}-1}< 1-\frac{2}{20^{10}-3}\)
=> A < B
a) Ta có: \(A=\frac{2^{2017}}{2^{2017}}+\frac{2^{2016}}{2^{2017}}+\frac{2^{2015}}{2^{2017}}+...+\frac{2^1}{2^{2017}}+\frac{1}{2^{2017}}\)
\(=\frac{1+2^1+2^2+...+2^{2016}+2^{2017}}{2^{2017}}\)
Đặt: B=\(1+2^1+2^2+...+2^{2017}\)
\(\Leftrightarrow2B=2^1+2^2+2^3+....+2^{2017}+2^{2018}\)
\(\Leftrightarrow2B-B=2^{2018}-1\)
\(\Leftrightarrow B=2^{2018}-1\)
\(\Rightarrow A=\frac{B}{2^{2017}}=\frac{2^{2018}-1}{2^{2017}}\)
Mik chỉ biết làm phần a thôi
b/ Sử dụng quy tắc: \(\frac{a+c}{b+c}< \frac{a}{b}\) với \(\left\{{}\begin{matrix}a;b;c>0\\a>b\end{matrix}\right.\)
\(B=\frac{2^{10}-1}{2^{10}-3}>\frac{2^{10}-1+2}{2^{10}-3+2}=\frac{2^{10}+1}{2^{10}-1}\)
\(\Rightarrow B>A\)
Mình làm luôn, koghi lại đề
A=\(\frac{5}{10}+\frac{5}{40}+\frac{5}{88}+\frac{5}{154}+\frac{5}{238}+\frac{5}{340}\)
A=\(\frac{5}{2.5}+\frac{5}{5.8}+\frac{5}{8.11}+\frac{5}{11.14}+\frac{5}{14.17}+\frac{5}{17.20}\)
A=\(\frac{5}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\right)\)
A=\(\frac{5}{3}.\left(\frac{1}{2}-\frac{1}{20}\right)\)
A=\(\frac{5}{3}.\frac{9}{20}\)
A=\(\frac{3}{4}\)
\(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1}{20^{10}-1}+\frac{2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3}{20^{10}-3}+\frac{2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
\(20^{10}-1>20^{10}-3\Rightarrow\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\)
=> A < B
Lời giải:
$A=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}=\frac{20^{10}-1}{20^{10}-3}=B$
Vậy $A< B$
A=20^10+1/20^10-1=1*2/20^10-1
B=20^10-1/20^10+3=1*2/20^10-3
vi 20^10-1>20^10-3
Suy ra 2/20^10-1<2/20^10-3
do \(\frac{10}{a}\)= \(\frac{20}{10}\)
=>ax20 = 10x10
=>ax20 = 100
=>a = 100: 20
=>a = 5
Vậy ...
a=(10.10):20=1