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
ta co A=\(\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=\frac{20^{10}-1}{20^{10}-1}+\frac{2}{20^{10}-1}\)=\(1+\frac{2}{20^{10}-1}\)
B=\(\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=\frac{20^{10}-3}{20^{10}-3}+\frac{2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
ta thấy\(20^{10}-1>20^{10}-3\) suy ra \(\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\)vậy A<B