Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Áp dụng bất đẳng thức :
\(\dfrac{a}{b}< \dfrac{a+m}{b+m}\)
Ta có :
\(A=\dfrac{10^{101}-1}{10^{102}-1}< \dfrac{10^{101}-1+11}{10^{102}-1+11}=\dfrac{10^{101}+10}{10^{102}+10}=\dfrac{10\left(10^{100}+1\right)}{10\left(10^{101}+1\right)}=\dfrac{10^{100}+1}{10^{101}+1}=B\)
\(\Leftrightarrow A< B\)
Ta có:
\(1-A=1-\dfrac{10^{101}-1}{10^{102}-1}=\dfrac{10^{102}-1\left(10^{101}-1\right)}{10^{102}-1}\) \(=\dfrac{10^{102}-1-10^{101}+1}{10^{102}-2}=\dfrac{10^{102}-10^{101}}{10^{102}-1}\)
\(=\dfrac{10^{101}\left(10-1\right)}{10^{101}\left(10-\dfrac{1}{10^{101}}\right)}=\dfrac{10-1}{10-\dfrac{1}{10^{101}}}=\dfrac{9}{10-\dfrac{1}{10^{101}}}\)\(\left(1\right)\)
\(1-B=1-\dfrac{10^{100}+1}{10^{101}+1}=\dfrac{10^{101}+1-\left(10^{100}+1\right)}{10^{101}+1}\)
\(=\dfrac{10^{101}+1-10^{100}-1}{10^{101}+1}\) \(=\dfrac{10^{101}-10^{100}}{10^{101}+1}=\dfrac{10^{100}\left(10-1\right)}{10^{100}\left(10+\dfrac{1}{10^{100}}\right)}\)
\(=\dfrac{10-1}{10+\dfrac{1}{10^{100}}}=\dfrac{9}{10+\dfrac{1}{100}}\)\(\left(2\right)\)
\(Từ\left(1\right);\left(2\right)\) \(=>A< B\)\(\left(đpcm\right)\)
CHÚC BẠN HỌC TỐT 
Ta có: \(A=\dfrac{1}{101^2}+\dfrac{1}{102^2}+\dfrac{1}{103^2}+\dfrac{1}{104^2}+\dfrac{1}{105^2}\)
\(A>\dfrac{1}{100.101}+\dfrac{1}{101.102}+\dfrac{1}{102.103}+\dfrac{1}{103.104}+\dfrac{1}{104.105}\)\(A>\dfrac{1}{100}-\dfrac{1}{101}+\dfrac{1}{101}-\dfrac{1}{102}+\dfrac{1}{102}-\dfrac{1}{103}+\dfrac{1}{103}-\dfrac{1}{104}+\dfrac{1}{104}-\dfrac{1}{105}\)\(A>\dfrac{1}{100}-\dfrac{1}{105}\)
\(A>\dfrac{1}{2100}\)
Mà \(B=\dfrac{1}{2^2.3.5^2.7}\)=\(\dfrac{1}{2100}\)
=> \(A>B\)
Vậy \(A>B\)
\(A=\sqrt{625}-\dfrac{1}{\sqrt{5}}=25-\dfrac{1}{\sqrt{5}}\)
\(B=\sqrt{576}-\dfrac{1}{\sqrt{6}}+1=24-\dfrac{1}{\sqrt{6}}+1=25-\dfrac{1}{\sqrt{6}}.\)
Vì \(\sqrt{5}< \sqrt{6}\) nên \(\dfrac{1}{\sqrt{5}}>\dfrac{1}{\sqrt{6}}.\)
Từ (1), (2) và (3) suy ra \(A< B.\)