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B=\(\frac{2011^{10}-1}{2011^{10}-3}\) <1 => \(\frac{2011^{10}-1}{2011^{10}-3}\) < \(\frac{2011^{10}-1+2}{2011^{10}-3+2}\) = \(\frac{2011^{10}+1}{2011^{10}-1}\) = A
=> B<A
ta co:B=2010-1/2010-3>1
=>B>2010-1+2/2010-3+2=2010+1/2010-1=A
vay A<B
để hỏi nhé !
A> \(\frac{10^n-2-2}{10^n-1-2}=\frac{10^n-4}{10^n-3}=B\)
=> A>B
\(A=\frac{10^8+2}{10^8-1}=\frac{\left(10^8-1\right)+3}{10^8-1}=\frac{10^8-1}{10^8-1}+\frac{3}{10^8-1}=1+\frac{3}{10^8-1}\)
\(B=\frac{10^8}{10^8-3}=\frac{\left(10^8-3\right)+3}{10^8-3}=\frac{10^8-3}{10^8-3}+\frac{3}{10^8-3}=1+\frac{3}{10^8-3}\)
Vì \(1+\frac{3}{10^8-1}<1+\frac{3}{10^8-3}\) nên A < B
Ta có :
A = 108 + 2 / 10 8 - 1 = 1 + 3 / 10 8 - 1
B = 108 / 10 8 - 3 = 1 + 3 / 108 - 3
Vì 3/ 108 - 1 < 3 / 108 - 3=> A < B
Chào bạn, bạn hãy theo dõi câu trả lời của mình nhé!
a) Ta có :
\(2^{225}=\left(2^3\right)^{75}=8^{75}\)
\(3^{151}=3^{150}\cdot3=\left(3^2\right)^{75}\cdot3=9^{75}\cdot3\)
Mà \(9^{75}>8^{75}=>9^{75}\cdot3>8^{75}=>3^{151}>2^{225}\)
b) Nhân cả vế A lẫn vế B với 102005, ta có :
\(10^{2005}A=-7+\frac{-15}{10}=\frac{-70}{10}+\frac{-15}{10}=\frac{-85}{10}\)
\(10^{2005}B=-15+\frac{-7}{10}=\frac{-150}{10}+\frac{-7}{10}=\frac{-157}{10}\)
Mà \(\frac{-85}{10}>\frac{-157}{10}=>10^{2005}A>10^{2005}B\)
\(=>A>B\)
Chúc bạn học tốt!
Ta có:
\(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
Ta lại có:
\(20^{10}-1>20^{10}-3\Rightarrow\frac{2}{2^{10}-1}< \frac{2}{2^{10}-3}\Rightarrow1+\frac{2}{2^{10}-1}< 1+\frac{2}{2^{10}-3}\)
Hay A<B
Ta có:
\(10A=10.\left(\frac{10^{234}+1}{10^{235}+1}\right)=\frac{10^{235}+10}{10^{235}+1}=\frac{10^{235}+1}{10^{235}+1}+\frac{9}{10^{235}+1}=1+\frac{9}{10^{235}+1}\)
\(10B=10.\left(\frac{10^{235}+1}{10^{236}+1}\right)=\frac{10^{236}+10}{10^{236}+1}=\frac{10^{236}+1}{10^{236}+1}+\frac{9}{10^{236}+1}=1+\frac{9}{10^{236}+1}\)
\(10^{235}+1<10^{236}+1\Rightarrow\frac{9}{10^{235}+1}\)\(>\)\(\frac{9}{10^{236}+1}\)
\(\Rightarrow1+\frac{9}{10^{235}+1}\)\(>\)\(1+\frac{9}{10^{236}+1}\)
\(\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
Vậy \(A>B\)