Ta thấy \(\frac{10^8+2}{10^8-1}=\frac{10^8-1+3}{10^8-1}=1+\frac{3}{10^8-1}\)

\(\frac{10^8}{10^8-3}=\frac{10^8-3+3}{10^8-3}=1+\frac{3}{10^8-3}\)

Vậy là em so sánh được rồi nhé :)

\(A=\frac{10^8+2}{10^8-1}=\frac{10^8-1+3}{10^8-1}=1+\frac{3}{10^8-1}\)

\(B=\frac{10^8}{10^8-3}=\frac{10^8-3+3}{10^8-3}=1+\frac{3}{10^8-3}\)

Vì \(\frac{3}{10^8-1}

\(\frac{10^8+2}{10^8-1}=1+\frac{3}{10^8-1}

 A < B đấy , chắc lun

\(A=\frac{10^8+2}{10^8-1}=\frac{\left(10^8-1\right)+3}{10^8-1}=\frac{10^8-1}{10^8-1}+\frac{3}{10^8-1}=1+\frac{3}{10^8-1}\)

\(B=\frac{10^8}{10^8-3}=\frac{\left(10^8-3\right)+3}{10^8-3}=\frac{10^8-3}{10^8-3}+\frac{3}{10^8-3}=1+\frac{3}{10^8-3}\)

Vì \(1+\frac{3}{10^8-1}<1+\frac{3}{10^8-3}\) nên A < B

Ta có : 

 A = 10⁸ + 2 / 10 ⁸ - 1 = 1 + 3 / 10 ⁸ - 1

B = 10⁸ / 10 ⁸ - 3 =  1 + 3 / 10⁸ - 3

Vì 3/ 10⁸ - 1 < 3 / 10⁸ - 3=> A < B

A = \(\dfrac{10^8+2}{10^8-1}\) = \(\dfrac{\left(10^8-1\right)+3}{10^8-1}\) = \(\dfrac{10^8-1}{10^8-1}\) + \(\dfrac{3}{10^8-1}\) = 1+ \(\dfrac{3}{10^8-1}\)

B = \(\dfrac{10^8}{10^8-3}\) = \(\dfrac{\left(10^8-3\right)+3}{10^8-3}\) = \(\dfrac{10^8-3}{10^8-3}\) +\(\dfrac{3}{10^8-3}\)= 1+ \(\dfrac{3}{10^8-3}\)

Vì 1 +\(\dfrac{3}{10^8-1}\) < 1 + \(\dfrac{3}{10^8-3}\) nên A < B

\(\dfrac{3}{10^8-3}\)\(\dfrac{3}{10^8-3}\)

Đề hình như sai rùi bn, ở A mẫu phải là 10⁸ - 1 chứ

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

Ta có:

\(B=\frac{10^8}{10^8-3}< \frac{10^8+2}{10^8-3+2}=\frac{10^8+2}{10^8-1}=A\)

=> B < A

trừ A cho 3/(10⁸-1)   (1)  = 1

trừ B cho 3/(10⁸-3)    (2) = 1

dễ thấy (1)>(2) suy ra A>B

\(A=\frac{10^8+2}{10^8+1}=1+\frac{1}{10^8+1}

10^8+2>10^8

10^8+1<10^^8-3

\(10^8=a\)

\(A-B=\frac{a+2}{a+1}-\frac{a}{a-3}=\frac{\left(a+2\right)\left(a-3\right)-a\left(a+1\right)}{\left(a+1\right)\left(a-3\right)}=\frac{-2a-6}{\left(a+1\right)\left(a-3\right)}