c) \(\frac{a+1}{a}<\frac{a+1+2}{a+2}=\frac{a+3}{a+2}\)    (áp dụng công thức \(\frac{a}{b}<\frac{a+m}{b+m}\))

\(\Rightarrow\frac{a+1}{a}<\frac{a+3}{a+2}\)

d) \(\frac{a}{a+6}<\frac{a+1}{a+6+1}=\frac{a+1}{a+7}\)

\(\Rightarrow\frac{a}{a+6}<\frac{a+1}{a+7}\)

\(3^{300}=\left(3^3\right)^{100}=27^{100}\)

\(5^{199}< 5^{200}\) mà \(5^{200}=25^{100}\)

\(25^{100}< 27^{100}\Rightarrow3^{300}>5^{200}>5^{199}\)

Trong hai phân số cùng tử nếu mẫu nào lớn hớn thì phân số đó bé hơn.

Vậy : \(\frac{1}{5^{199}}>\frac{1}{3^{300}}\)

1/5^99 > 1 / 3^300

2.  a) \(3^{200}=\left(3^2\right)^{100}=9^{100}\)

          \(2^{300}=\left(2^3\right)^{100}=8^{100}\)

Vì \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)

b) \(71^{50}=\left(71^2\right)^{25}=5041^{25}\)

     \(37^{75}=\left(3^3\right)^{25}=27^{25}\)

Vì \(5041^{25}>27^{25}\Rightarrow71^{50}>37^{75}\)

c) \(\frac{201201}{202202}=\frac{201201:1001}{202202:1001}=\frac{201}{202}\)

      \(\frac{201201201}{202202202}=\frac{201201201:1001001}{202202202:1001001}=\frac{201}{202}\)

Vì \(\frac{201}{202}=\frac{201}{202}\Rightarrow\frac{201201}{202202}=\frac{201201201}{202202202}\)

Gyvyghghgbhg

\(A=\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2010}\)

\(A=\frac{4064340600}{4066362660}+\frac{4064341605}{4066362660}+\frac{4070408792}{4066362660}\)

\(A=3,000000742\)

\(B=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+....+\frac{1}{17}\)

\(B=1,939552553\)

vì đây là so sánh hai dòng phân số nên ta  đổi ra thập phân nhé

do 3,000000742 > 1,939552553 và 3 > 1 Nên A > B nhé

đúng thì k nhé

chúc học giỏi !!!!

 A > B nha bạn !!! ( $ _ $ )%%%

\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)

\(\text{B = }\text{ }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)

\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)

\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)

=> A > B

Vậy A > B