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 :
\(25^{1008}=\left(5^2\right)^{1008}=5^{2.1008}=5^{2016}\)
mà \(5^{2017}>5^{2016}\)
\(\Rightarrow\)\(5^{2017}>\left(5^2\right)^{1008}\)
\(\Rightarrow\)\(5^{2017}>25^{1008}\)
có \(5^{2017}=\left(5^2\right)^{1008}\times5\)\(=25^{1008}\times5\)
mà \(=25^{1008}\times5\)> \(25^{1008}\)
nên \(5^{2017}>25^{1008}\)
So sánh A=\(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{2021}\)và B=20. So sánh A và B
\(A=\dfrac{10^{99}+1}{10^{100}+1}\)
\(\Leftrightarrow10A=\dfrac{10\left(10^{99}+1\right)}{10^{100}+1}\)
\(\Leftrightarrow10A=\dfrac{10^{100}+10}{10^{100}+1}=\dfrac{10^{100}+1+9}{10^{100}+1}=1+\dfrac{9}{10^{100}+1}\)
\(B=\dfrac{10^{100}+1}{10^{101}+1}\)
\(\Leftrightarrow10B=\dfrac{10\left(10^{100}+1\right)}{10^{101}+1}\)
\(\Leftrightarrow10B=\dfrac{10^{101}+10}{10^{101}+1}=\dfrac{10^{101}+1+9}{10^{101}+1}=1+\dfrac{9}{10^{101}+1}\)
Do \(\dfrac{9}{10^{100}+1}>\dfrac{9}{10^{101}+1}\) nên \(10A>10B\)
\(\Rightarrow A>B\)
Áp dụng tính chất:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(B=\dfrac{10^{100}+1}{10^{101}+1}< 1\)
\(B< \dfrac{10^{100}+1+9}{10^{101}+1+9}\)
\(B< \dfrac{10^{100}+10}{10^{101}+10}\)
\(B< \dfrac{10\left(10^{99}+1\right)}{10\left(10^{100}+1\right)}\)
\(B< \dfrac{10^{99}+1}{10^{100}+1}=A\)
\(B< A\)
A=
\(\dfrac{10^{101}-1}{10^{102}-1}< \dfrac{10^{101}-1+11}{10^{102}-1+11}=\dfrac{10^{101}+10}{10^{102}+10}=\dfrac{10\left(10^{100}+1\right)}{10\left(10^{101}+1\right)}=B\)
Vậy A<B
CHÚC BẠN HỌC TỐT
Ta có :
\(A=\dfrac{10^{11}-1}{10^{12}-1}< 1\)
\(\Leftrightarrow A< \dfrac{10^{11}-1+11}{10^{12}-1+11}=\dfrac{10^{11}+10}{10^{12}+10}=\dfrac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}=\dfrac{10^{10}+1}{10^{11}+1}=B\)
Vậy \(\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{10^{10}+1}{10^{11}+1}\)
Vậy...
a: \(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\)
\(=-\dfrac{1}{10}\)
9<10
=>1/9>1/10
=>\(-\dfrac{1}{9}< -\dfrac{1}{10}\)
=>\(A>-\dfrac{1}{9}\)
b: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{10}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{11}{10}\)
\(=\dfrac{-1}{10}\cdot\dfrac{11}{2}=\dfrac{-11}{20}\)
20<21
=>\(\dfrac{11}{20}>\dfrac{11}{21}\)
=>\(-\dfrac{11}{20}< -\dfrac{11}{21}\)
=>\(B< -\dfrac{11}{21}\)
Ta có:
10A=10^102-10/10^102-1
10A=1-9/10^102-1
10B=10^101+10/10^101+1
10B=1+9/10^101+1
suy ra 10B>10A
Vậy B>A