Question

Đúng cho 3k

Answer 1

Vậy mà cũng đăng

Answer 2

C1: Áp dụng khi \(\dfrac{a}{b}>1\) thì \(\dfrac{a}{b}>\dfrac{a+m}{b+m}\left(m\in N^{\circledast}\right)\)

Ta thấy \(\dfrac{20^{2015}-1}{20^{2015}-3}>1\)

\(\Rightarrow B=\dfrac{20^{2015}-1}{20^{2015}-3}>\dfrac{20^{2015}-1+2}{20^{2015}-3+2}=\dfrac{20^{2015}+1}{20^{2015}-1}=A\)

Vậy \(B>A\)

Answer 3

C2:

\(A=\dfrac{20^{2015}+1}{20^{2015}-1}=\dfrac{20^{2015}-1+2}{20^{2015}-1}=\dfrac{20^{2015}-1}{20^{2015}-1}+\dfrac{2}{20^{2015}-1}=1+\dfrac{2}{20^{2015}-1}\)\(B=\dfrac{20^{2015}-1}{20^{2015}-3}=\dfrac{20^{2015}-3+2}{20^{2015}-3}=\dfrac{20^{2015}-3}{20^{2015}-3}+\dfrac{2}{20^{2015}-3}=1+\dfrac{2}{20^{2015}-3}\)\(20^{2015}-1>20^{2015}-3\\ \Rightarrow\dfrac{2}{20^{2015}-1}< \dfrac{2}{20^{2015}-3}\\ \Rightarrow1+\dfrac{2}{20^{2015}-1}< 1+\dfrac{2}{20^{2015}-3}\\ \Rightarrow A< B\)

Answer 4

Cả 2 cách đều ghi thêm kết luận nha, mk quên ghi rồi

Vậy \(A< B\)

Answer 5

Ta có:

Nếu:

\(\dfrac{a}{b}>1\Rightarrow\dfrac{a+m}{b+m}>1\left(m\in N\cdot\right)\)

B=\(\dfrac{20^{2015}-1}{20^{2015}-3}>1\)

B=\(\dfrac{20^{2015}-1}{20^{2015}-3}>\)\(\dfrac{20^{2015}-1+2}{20^{2015}-3+2}\)\(>\dfrac{20^{2015}+1}{20^{2015}-1}=A\)

Vậy B>A

Answer 6

Tui làm trên kia rồi kìa