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

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

\(\Rightarrow1+\frac{2}{20^{10}-1}<1+\frac{2}{20^{10}-3}\)

=>A<B

Theo đề, ta có:

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

Suy ra  \(B<\frac{20^{10}+1}{20^{10}-1}\)

Mà \(A=\frac{20^{10}+1}{20^{10}-1}\)

Nên ​​​B < A 

Giải:

Ta có:

A=20¹⁰+1/20¹⁰-1

A=20¹⁰-1+2/20¹⁰-1

A=1+2/20¹⁰-1

Tương tự:

B=20¹⁰-1/20¹⁰-3

B=20¹⁰-3+2/20¹⁰-3

B=1+2/20¹⁰-3

Vì 2/20¹⁰-1<2/20¹⁰-3 nên A<B

Chúc bạn học tốt!

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

Nên A = B

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

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

Vì \(\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\Rightarrow1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}\Rightarrow A< B\)

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

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

Do 20¹⁰ - 1 > 20¹⁰ - 3

\(\Rightarrow\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\Rightarrow1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}\)

Hay A < B 

Ai thấy tớ đúng k nha

Ta thấy B=20^10-1/20^10-3 là phân số lớn hơn 1.

Theo tính chất nếu a/b>1 thì a/b > a+n/b+n ( n khác 0 )

Ta có : 20^10-1/20^10-3 > 20^10-1+2/20^10-3+2

          <=> B > 20^10+1/20^10-3 = A

          <=> B > A

          Vậy B > A    

Lời giải:

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

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

Vì $20^{10}-1> 20^{10}-3$

$\Rightarrow \frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}$

$\Rightarrow 1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}$

$\Rightarrow A< B$

Áp dụng a/b > 1 => a/b > a+m/b+m (a;b;m thuộc N*)

Ta có:

B = 20¹⁰ - 1/20¹⁰ - 3 > 20¹⁰ - 1 + 2/20¹⁰ - 3 + 2

=> B > 20¹⁰ + 1/20¹⁰ - 1 = A

=> B > A

Áp dụng a/b > 1 => a/b > a+m/b+m (a;b;m thuộc N*)

Ta có:

B = 20¹⁰ - 1/20¹⁰ - 3 > 20¹⁰ - 1 + 2/20¹⁰ - 3 + 2

=> B > 20¹⁰ + 1/20¹⁰ - 1 = A

=> B > A

 A = (20^10 + 1)/(20^10 - 1) = 1 - 2/(20^10 - 1) 
B = (20^10 - 1)/(20^10 - 3) = 1 - 2/(20^10 - 3) 
Do cái 20^10 - 1 lớn hơn nên 2/(20^10 - 3) lớn hơn 2/(20^10 - 1) => A > B

A<B

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