Question

So sánh :

\(A=\frac{20^{10}+1}{20^{10}-1}vàB=\frac{20^{10-1}}{20^{10}-3}\)

Answer 1

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

Answer 2

A<B

Answer 3

A>B

 

Answer 4

Ta có: A=20^10+1/20^10-1

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

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

A=1+2/20^10-1

Ta có:B=20^10-1/20^10-3

          B=20^10-3+2/20^10-3

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

           B=1+2/20^10-3

 Vì 2/20^10-3>2/20^10-1

Nên 1+2/20^10-3>1+2/20^10-1

Hay          B>A

Vậy A<B

Answer 5

Ta có:

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

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

Vì \(\dfrac{2}{20^{10}-1}>\dfrac{2}{20^{10}-3}\)

\(\Rightarrow\dfrac{20^{10}+1}{20^{10}-1}< \dfrac{20^{10}-1}{20^{10}-3}\)

\(\Rightarrow A< B\)

Answer 6

nhân với 10