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

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

Vì 10⁸-1 > 10⁸-3

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

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

Ta có: A = 10⁸ + 2/ 10⁸ - 1 = 3/10⁸ - 1                           

              B = 10⁸ / 10⁸ - 3 = 3 / 10⁸ -3             

Vì 3 / 10⁸ - 1 < 3 / 10⁸ -3 nên           

Nên A< B

A=10^8+2/10^8-1=10^8-1+3/10^8-1 
=10^8-1/10^8-1+3/10^8-1=1+3/10^8-1=1/3/... 
B=10^8/10^8-3=10^8-3+3/10^8-3 
=10^8-3/10^8-3+3/10^8-3=1+3/10^8-3=1/3/... 
tu (1) va (2) =>1/3/10^8-1<1/3/10^8-3(vi phân so nao 
co mau be hon thi phân so do lon hon nen 10^8-1>10^8-3) 
Hay A<B 

k mk nha

Mình làm câu a) nha!!!

+) \(A=2009^{2010}+2009^{2009}\)

        \(=2009^{2009}.\left(2009+1\right)\)

        \(=2009^{2009}.2010\)

+) \(B=2010^{2010}=2010^{2009}.2010\)

Vì \(2010^{2009}>2009^{2009}\)nên \(2010^{2009}.2010>2009^{2009}.2010\)hay \(B>A\)

Vậy \(A< B\)

Hok tốt nha^^

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

vậy A>B

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

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

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

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

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

Vì \(\frac{3}{10^8-1}>\frac{3}{10^8-3}\Rightarrow A>B\)

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

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

Nhận thầy 10⁸ - 1 > 10⁸ - 3

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

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

=> A < B

b) 17C = \(\frac{17\left(17^{203}+1\right)}{17^{204}+1}=\frac{17^{204}+1+16}{17^{204}+1}=1+\frac{16}{17^{204}+1}\)

17D = \(\frac{17\left(17^{202}+1\right)}{17^{203}+1}=\frac{17^{203}+1+16}{17^{203}+1}=1+\frac{16}{17^{203}+1}\)

Nhận thầy 17²⁰³ + 1 < 17²⁰⁴ + 1

=> \(\frac{16}{17^{203}+1}>\frac{16}{17^{204}+1}\)

=> \(\frac{16}{17^{203}+1}+1>\frac{16}{17^{204}+1}+1\Rightarrow17C>17D\Rightarrow C>D\) 

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

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

Ta có: \(\frac{3}{10^8-1}< \frac{3}{10^8-3}\)

\(\Rightarrow1+\frac{3}{10^8-1}< 1+\frac{3}{10^8-3}\)

\(\Rightarrow\frac{10^8+2}{10^8-1}< \frac{10^8}{10^8-3}\)

Ta có:\(\frac{196}{197}+\frac{197}{198}=\left(1-\frac{1}{197}\right)+\left(1-\frac{1}{198}\right)=2-\frac{1}{197}-\frac{1}{198}>2-1=1\)

Mà \(\frac{196+197}{197+198}< \frac{197+198}{197+198}=1\)

\(\Rightarrow\frac{196}{197}+\frac{197}{198}>\frac{196+197}{197+198}\)

c) tương tự câu a

Tham khảo nhé~