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

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

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

=> A < B

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>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

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

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 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=(20^10+1/20^10-1)=(20^10-1+2/20^10-1)=(20^10-1/20^10-1)+(2/20^10-1)=1+(2/20^10-1).

Tương tự ,B=1+(2/20^10-3).
Vì 2/20^10-1>2/20^10-3(vì 20^10-1>20^10-3).

=>A<B.

Vậy A<B.

tk mk nha đúng 1000000%,2b bạn kia sai rùi