Question

So sánh: A = 20^10+1/20^10-1 và B = 20^10-1/20^10-3

Ai nhanh mk tích

Answer 1

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

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

Từ đó \(\Rightarrow\) A < B 

\(hnhaminhhlai\)

Answer 2

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

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

vì 20¹⁰-1>20¹⁰-3

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

=>A<B

Answer 3

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

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

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

Suy ra: A < B

Answer 4

Nobita Kun sai rồi \(B=\frac{2010-1}{2010-3}=\frac{2010-3+2}{2010-3}=1+\frac{2}{2010-3}\)