Vì \(20^{10}-1>20^{10}-3\)

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

vậy \(A< B\)

Ta có:

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

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

Vậy A và B bằng nhau

Tính A và B rồi ta so sánh:

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

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

Mà \(1\) = \(1\)

Nên: A = B

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

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

\(20^{10}-1>20^{10}-3\Rightarrow\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\)

=> A < B

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

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

Vì \(\frac{2}{2^{10}-1}

ta co:B=20¹⁰-1/20¹⁰-3>1

=>B>20¹⁰-1+2/20¹⁰-3+2=20¹⁰+1/20¹⁰-1=A

vay A<B

Vì \(20^{10}-1>20^{10}-3\)

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

\(\Rightarrow A< B\)
 

Ta có : \(A=\frac{20^{10}+1}{20^{10}-1}=\frac{\left(20^{10}-1\right)+2}{20^{10}-1}\)

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

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

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

Do : \(20^{10}-1>20^{10}-3\)

\(\Rightarrow\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\Rightarrow1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}\)

Vậy : \(A< B\)

Lời giải:

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

Vậy $A< B$

Ta thấy B=20^10-1/20^10-3 là phân số lớn hơn 1.

Theo tính chất nếu a/b>1 thì a/b > a+n/b+n ( n khác 0 )

Ta có : 20^10-1/20^10-3 > 20^10-1+2/20^10-3+2

          <=> B > 20^10+1/20^10-3 = A

          <=> B > A

          Vậy B > A    

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

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

Do \(\frac{2}{20^{10}-1}>\frac{2}{20^{10}-3}\) nên \(\frac{20^{10}+1}{20^{10}-1}