Lời giải:

a) Xét hiệu \(\frac{a+n}{b+n}-\frac{a}{b}=\frac{(a+n).b-a(b+n)}{b(b+n)}=\frac{n(b-a)}{b(b+n)}\)

Nếu $b>a$ thì $\frac{a+n}{b+n}-\frac{a}{b}>0\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$

Nếu $b<a$ thì $\frac{a+n}{b+n}-\frac{a}{b}<0\Rightarrow \frac{a+n}{b+n}<\frac{a}{b}$

Nếu $b=a$ thì $\frac{a+n}{b+n}-\frac{a}{b}=0\Rightarrow \frac{a+n}{b+n}=\frac{a}{b}$

b) Rõ ràng $10^{11}-1< 10^{12}-1$. 

Đặt $10^{11}-1=a; 10^{12}-1=b; 11=n$ thì: $a< b$; $A=\frac{a}{b}$ và $B=\frac{10^{11}+10}{10^{12}+10}=\frac{a+n}{b+n}$

Áp dụng kết quả phần a:

$b>a\Rightarrow \frac{a+n}{b+n}>\frac{a}{b}$ hay $B>A$

Ta có :\(a=\dfrac{10^{11}-1}{10^{12}-1}\Rightarrow10a=\dfrac{10^{12}-10}{10^{12}-1}=\dfrac{10^{12}-1-9}{10^{12}-1}=1-\dfrac{9}{10^{12}-1}\)

\(b=\dfrac{10^{10}+1}{10^{11}+1}\Rightarrow10b=\dfrac{10^{11}+10}{10^{11}+1}=\dfrac{10^{11}+1+9}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\)

Ta có : \(1-\dfrac{9}{10^{12}-1}\le1+\dfrac{9}{10^{11}+1}\) hay \(10a< 10b\Rightarrow a< b\)

Nếu:

\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

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

\(A< \dfrac{10^{11}-1+11}{10^{12}-1+11}\Rightarrow A< \dfrac{10^{11}+10}{10^{12}+10}\Rightarrow A< \dfrac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}\Rightarrow A< \dfrac{10^{10}+1}{10^{11}+1}=B\)

\(\Rightarrow A< B\)

ta có :

\(A=\dfrac{10^{11}-1}{10^{12}-1}\\ 10A=\dfrac{10^{12}-10}{10^{12}-1}=1-\dfrac{9}{10^{12}-1}\\ =>10A< 1\\ B=\dfrac{10^{10}+1}{10^{11}+1}\\ 10B=\dfrac{10^{11}+10}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\\ =>10B>1\)

=> 10A<10B =>A<B

vậy A bé hơn B

Ta có: \(\dfrac{10^{11}-1}{10^{12}-1}< \dfrac{10^{11}-1+11}{10^{12}-1+11}\)

\(\Rightarrow A< \dfrac{10^{11}+10}{10^{12}+10}\)

\(\Rightarrow A< \dfrac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}\)

\(\Rightarrow A< \dfrac{10^{10}+1}{10^{11}+1}\)

\(\Rightarrow A< B\)

Vậy \(A< B\).

Cách 2:

Ta có: \(10A=\dfrac{10^{12}-10}{10^{12}-1}=1-\dfrac{9}{10^{12}-1}\)

\(10B=\dfrac{10^{11}+10}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\)

Vì \(\dfrac{9}{10^{12}-1}< \dfrac{9}{10^{11}+1}\Rightarrow1-\dfrac{9}{10^{12}-1}< 1+\dfrac{9}{10^{11}+1}\)

\(\Rightarrow10A< 10B\Rightarrow A< B\)

Vậy A < B

Cảm ơn hỏi giùm!

Lời giải:
a.

\(\frac{n+1}{n+2}=\frac{n+1}{n+2}+1-1=\frac{2n+3}{n+2}-1\)

\(> \frac{2n+3}{n+3}-1=\frac{(n+3)+n}{n+3}-1=\frac{n}{n+3}\)

b.

\(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{(10^{12}-1)-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}<1\)

\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{(10^{11}+1)+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)

$\Rightarrow 10A< 10B\Rightarrow A< B$

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...

Vì \(10^{11}-1< 10^{12}-1\)

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

ta có: \(A=\dfrac{10.10^{10}-1}{10.10^{11}-1}=\dfrac{10^{10}-1}{10^{11}-1}\)

so sánh: \(A=\dfrac{10^{10}-1}{10^{11}-1}\)và \(B=\dfrac{10^{10}+1}{10^{11}+1}\)

\(\Rightarrow A< B\)

10A=\(\dfrac{10^{12}-10}{10^{12}-1}=1-\dfrac{9}{10^{12}-1}\)

10B =\(\dfrac{10^{11}+10}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\)

\(\dfrac{9}{10^{12}-1}< \dfrac{9}{10^{12}+1}\) =>\(1-\dfrac{9}{10^{12}-1}< 1+\dfrac{9}{10^{11}+1}\)

=> 10A < 10B

=> A<B

Vậy A < B

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

\(A=\dfrac{10.10^{10}-1}{10.10^{11}-1}\)và \(B=\dfrac{10^{10}+1}{10^{11}+1}\)

Vậy A<B