ko bíb

Có : 10A = 10.(10^11-1)/10^12-1 = 10^12-10/10^12-1 

Vì : 0 < 10^12-10 < 10^12-1 => 10A < 1 (1)

10B = 10.(10^10+1)/10^11+1 = 10^11+10/10^11+1

Vì : 10^11+10 > 10^11+1 > 0 => 10B > 1 (2)

Từ (1) và (2) => 10A < 10B

=> A < B

Tk mk nha

\(A=\frac{10^{11}-1}{10^{12}-1}\)

\(B=\frac{10^{10}+1}{10^{11}+1}\)

Mà \(\frac{10^{11}-1}{10^{12}-1}< 1\); \(\frac{10^{10}+1}{10^{11}+1}< 1\)

\(\Rightarrow\)\(A,B< 1\)

Ta có:

\(10^{11}-1>10^{10}+1\); \(10^{12}-1>10^{11}+1\)

\(\Rightarrow A>B\)

Vậy A > B

a) Ta có

A = n / n+1 = 1-(1/n+1)

A = n+2 / n+3 = 1-(1/n+3)

Vì 1/n+1 > 1/n+3

=> n/n+1 < n+2/n+3 

=> A<B

a: Ta có: \(A=\frac59+\left(-\frac57\right)+\left(-\frac{20}{48}\right)+\frac{8}{12}+\left(-\frac{21}{48}\right)\)

\(=\frac59-\frac57-\frac{41}{48}+\frac{32}{48}\)

\(=\frac{35-45}{63}-\frac{9}{48}=\frac{-10}{63}-\frac{3}{16}=\frac{-160-189}{63\cdot16}=\frac{-349}{1008}\)

b: \(B=\left(-\frac59\right)+\frac{8}{15}+\left(-\frac{2}{11}\right)+\left(\frac{4}{-9}\right)+\frac{2}{45}\)

\(=\left(-\frac59-\frac49\right)+\frac{8}{15}+\frac{2}{45}-\frac{2}{11}\)

\(=-1-\frac{2}{11}+\frac{24}{45}+\frac{2}{45}=-\frac{13}{11}+\frac{26}{45}=\frac{-13\cdot45+26\cdot11}{11\cdot45}=\frac{-299}{495}\)

c: \(\frac{1}{11}>\frac{1}{20};\frac{1}{12}>\frac{1}{20};\ldots;\frac{1}{20}=\frac{1}{20}\)

Do đó: \(\frac{1}{11}+\frac{1}{12}+\cdots+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+\cdots+\frac{1}{20}\)

=>S>10/20

=>S>1/2

\(A=\dfrac{10^{11}+1}{10^{12}-1}\)

\(\Rightarrow10A=\dfrac{10^{11}+1}{10^{12}-1}.10\)

\(\Rightarrow10A=\dfrac{10\left(10^{11}+1\right)}{10^{12}-1}\)

\(\Rightarrow10A=\dfrac{10^{12}-10}{10^{12}-1}\)

\(B=\dfrac{10^{10}+1}{10^{11}+1}\)

\(\Rightarrow10B=\dfrac{10^{10}+1}{10^{11}+1}.10\)

\(\Rightarrow10B=\dfrac{\left(10^{10}+1\right).10}{10^{11}+1}\)

\(\Rightarrow10B=\dfrac{10^{11}+10}{10^{11}+1}\)

Ta thấy:

 \(10^{12}-1>10^{12}-10>0\Rightarrow10A< 1\)

\(0< 10^{11}+1< 10^{11}+10\Rightarrow10B>1\)

Mà \(10A< 1;10B>1\)

\(\Rightarrow B>A\).