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 $20^{10}-1> 20^{10}-3$

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

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

$\Rightarrow A< B$

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!

ta có: 20¹⁰ + 1 > 20¹⁰ - 1

\(\Rightarrow A=\frac{20^{10}+1}{20^{10}-1}>1\)

Lại có: 20¹⁰ -1 < 20¹⁰ - 3

\(\Rightarrow B=\frac{20^{10}-1}{20^{10}-3}< 1\)

=> A > B

\(A=\frac{2010+1}{2010-1}\)

\(A=1+\frac{2}{2010-1}>1\)

\(B=\frac{2010-1}{2010-3}\)

\(B=1-\frac{2}{2010-3}<1\)

Từ đó A > B

Ta thấy : A =\(\frac{20^{10}+1}{20^{10}-1}>1\) 
Ta có : A=\(\frac{20^{10}+1}{20^{10}-1}>\frac{20^{10}+1-2}{20^{10}-1-2}=\frac{20^{10}-1}{20^{10}-3}=B\)
Vậy A > B

cám ơn bạn

Theo đề ta có:

A= 20^10+1/20^10-1 (1)

Từ (1) ta có: 20^10+1/20^10-1>20^10+1-2/20^10-1-2=20^10-1/20^10-3=B.

Vậy A>B

Ta thấy:\(A=\frac{20^{10}+1}{20^{10}-1}>1\)

Ta có: \(A=\frac{20^{10}+1}{20^{10}-1}>\frac{20^{10}+1-2}{20^{10}-1-2}=\frac{20^{10}-1}{20^{10}-3}=B\)

Vậy \(A>B\)

Ta có:

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

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

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

\(B=\frac{20^{10}-1}{20^{10}-3}\)

\(=\frac{20^{10}-3+2}{20^{10}-3}\)

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

Ta lại có:

\(20^{10}-1>20^{10}-3\)

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

\(\Rightarrow\)\(1+\frac{2}{2^{10}-1}< 1+\frac{2}{2^{10}-3}\)

Vậy ta kết luận A < B

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

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

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