Giải:

a) A=17¹⁸+1/17¹⁹+1

17A=17¹⁹+17/17¹⁹+1

17A=17¹⁹+1+16/17¹⁹+1

17A=1+16/17¹⁹+1

Tương tự:

B=17¹⁷+1/17¹⁸+1

17B=17¹⁸+17/17¹⁸+1

17B=17¹⁸+1+16/17¹⁸+1

17B=1+16/17¹⁸+1

Vì 16/17¹⁹+1<16/17¹⁸+1 nên 17A<17B

⇒A<B

b) A=10⁸-2/10⁸+2

    A=10⁸+2-4/10⁸+2

    A=1+-4/10⁸+2

Tương tự:

B=10⁸/10⁸+4

B=10⁸+4-4/10⁸+1

B=1+-4/10⁸+1

Vì -4/10⁸+2>-4/10⁸+1 nên A>B

c)A=20¹⁰+1/20¹⁰-1

   A=20¹⁰-1+2/20¹⁰-1

   A=1+2/20¹⁰-1

Tương tự:

B=20¹⁰-1/20¹⁰-3

B=20¹⁰-3+2/20¹⁰-3

B=1+2/20¹⁰-3

Vì 2/20¹⁰-3>2/20¹⁰-1 nên B>A

⇒A<B

Chúc bạn học tốt!

a, Ta có : \(10^{15}\cdot11=10^{15}\left(10+1\right)=10^{16}+10^{15}\)

Vì \(10^{16}+10^{15}>10^{16}+10\)

\(\Rightarrow\dfrac{10^{16}+10^{15}}{10^{16}+1}>\dfrac{10^{16}+10}{10^{16}+1}\)

Hay A>B

b, Ta có : \(C=\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}}{10^{10}-1}+\dfrac{1}{10^{10}-1}\)

\(D=\dfrac{10^{10}-1}{10^{13}-3}=\dfrac{10^{10}}{10^{13}-3}+\dfrac{-1}{10^{13}-3}\)

Vì \(\dfrac{10^{10}}{10^{10}-1}>\dfrac{10^{10}}{10^{13}-3};\dfrac{1}{10^{10}-1}>\dfrac{-1}{10^{13}-3}\)

\(\Rightarrow\dfrac{10^{10}+1}{10^{10}-1}>\dfrac{10^{10}-1}{10^{13}-3}\)

Hay C > D

Giải:

Ta có:

A=20¹⁰+1/20¹⁰-1

A=20¹⁰-1+2/20¹⁰-1

A=1+2/20¹⁰-1

Tương tự:

B=20¹⁰-1/20¹⁰-3

B=20¹⁰-3+2/20¹⁰-3

B=1+2/20¹⁰-3

Vì 2/20¹⁰-1<2/20¹⁰-3 nên A<B

Chúc bạn học tốt!

Lời giải:

$A=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}$

$B=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}$

Vì $20^{10}-1> 20^{10}-3$

$\Rightarrow \frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}$

$\Rightarrow 1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}$

$\Rightarrow A< B$

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.

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

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

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

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:

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

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

\(A=\dfrac{20^{10}+1}{20^{10}-1}=\dfrac{20^{10}-1}{20^{10}-1}+\dfrac{2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)

\(B=\dfrac{20^{10}-1}{20^{10}-3}=\dfrac{20^{10}-3}{20^{10}-3}+\dfrac{2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\)

\(\dfrac{2}{20^{10}-1}>\dfrac{2}{20^{10}-3}\Leftrightarrow A>B\)

Ta có: \(A=\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}-1+2}{10^{10}-1}=1+\dfrac{2}{10^{10}-1}\)

\(B=\dfrac{10^{10}-1}{10^{10}-3}=\dfrac{10^{10}-3+2}{10^{10}-3}=1+\dfrac{2}{10^{10}-3}\)

Vì \(\dfrac{2}{10^{10}-1}< \dfrac{2}{10^{10}-3}\Rightarrow1+\dfrac{2}{10^{10}-1}< 1+\dfrac{2}{10^{10}-3}\)

\(\Rightarrow A< B\)

Vậy A < B

\(A=\dfrac{10^{10}+1}{10^{10}-1}=\dfrac{10^{10}-1+2}{10^{10}-1}=1+\dfrac{2}{10^{10}-1}\)

\(B=\dfrac{10^{10}-1}{10^{10}-3}=\dfrac{10^{10}-3+2}{10^{10}-3}=1+\dfrac{2}{10^{10}-3}\)

Vì \(10^{10}-1>10^{10}-3\) nên ta có

\(\dfrac{2}{10^{10}-1}< \dfrac{2}{10^{10}-3}\)

Vậy \(A< B\)

B = \(\dfrac{10^{10}-1}{10^{10}-3}\)<\(\dfrac{10^{10}-1+2}{10^{10}-3+2}\)=\(\dfrac{10^{10}+1}{10^{10}-1}\)= A

Vậy A > B