Lời giải:

$B=\frac{10^{11}+10}{10^{12}+10}$

Đặt $10^{11}-1=a; 10^{12}-1=b$ thì $0< a< b$. Khi đó:

$A-B=\frac{a}{b}-\frac{a+11}{b+11}=\frac{11(a-b)}{b(b+11)}<0$

$\Rightarrow A< B$

Dễ vãi

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 $B>A$

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 :

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

=> 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!

a: \(\dfrac{15}{8}-\dfrac{13}{8}=\dfrac{15-13}{8}=\dfrac{2}{8}=\dfrac{1}{4}\)

b: \(\dfrac{7}{15}-\dfrac{2}{15}=\dfrac{7-2}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)

c: \(\dfrac{11}{12}-\dfrac{2}{12}=\dfrac{11-2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

d: \(\dfrac{19}{7}-\dfrac{5}{7}=\dfrac{19-5}{7}=\dfrac{14}{7}=2\)

`1/8 -1/2 + (-11/12 + 1)`

`=1/8 -1/2 + (-11/12 +12/12)`

`=1/8 -1/2 + 1/12`

`= 1/8- 4/8+1/12`

`= -3/8 + 1/12`

`=-7/24`

`---------`

`3/5 -(-7/10) + (-13/10)`

`= 3/5 + 7/10 + (-13/10)`

`= 6/10 + 7/10 + (-13/10)`

`= 13/10 +(-13/10)`

`= 0/10=0`

\(\dfrac{1}{8}-\dfrac{1}{2}+\left(\dfrac{-11}{12}+1\right)\\ =\dfrac{-3}{8}+\dfrac{1}{12}\\ =\dfrac{-7}{24}\\ \dfrac{3}{5}-\dfrac{-7}{10}+\left(-\dfrac{13}{10}\right)\\ =\dfrac{13}{10}-\dfrac{13}{10}\\ =0\)

\(\dfrac{1}{8}-\dfrac{1}{2}+\left(\dfrac{-11}{12}+1\right)=\dfrac{-3}{8}+\dfrac{1}{12}=\dfrac{-7}{24}\)

\(\dfrac{3}{5}-\dfrac{-7}{10}+\dfrac{-13}{10}=\dfrac{6}{10}-\dfrac{-7}{10}+\dfrac{-13}{10}=\dfrac{13}{10}+\dfrac{-13}{10}=\dfrac{0}{10}=0\)

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 \(10^{11}-1< 10^{12}-1\)

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