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:

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

Hay A<B

A<B

A>B

ta co:B=20¹⁰-1/20¹⁰-3>1

=>B>20¹⁰-1+2/20¹⁰-3+2=20¹⁰+1/20¹⁰-1=A

vay A<B

A=20^10+1/20^10-1=1*2/20^10-1

B=20^10-1/20^10+3=1*2/20^10-3

vi 20^10-1>20^10-3

Suy ra 2/20^10-1<2/20^10-3

ta thấy B>1 nên B=\(\frac{20^{10}-1}{20^{10}-3}\)>\(\frac{20^{10}-1+2}{20^{100}-3+2}\)=\(\frac{20^{10}+1}{20^{10}-1}\)=A

vậy B>A

nếu ko hiểu thì tham khảo trong SBT lớp 6 bài so sánh PS ấy

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 co B>1

=>A=20^10+1​'20^10-1<20^10+1/2010-3+1

=>A<20^10+1/20^10-3+1

=>A<20^10-1/20^10-3

=>A<B

Vậy A<B

So sánh \(\left(20^{10}+1\right)^2\)và \(\left(20^{10}-1\right)^2\)

\(20^{10}-1< 20^{10}+1\)

\(\Leftrightarrow\left(20^{10}+1\right)^2>\left(20^{10}-1\right)^2\)

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