Ta có : 

\(1-\frac{38}{133}=\frac{133}{133}-\frac{38}{133}=\frac{95}{133}=\frac{5}{7}\)

\(1-\frac{129}{344}=\frac{344}{344}-\frac{129}{344}=\frac{215}{344}=\frac{5}{8}\)

Vì \(\frac{5}{7}>\frac{5}{8}\) nên \(\frac{38}{133}< \frac{129}{344}\)

\(\Rightarrow\)\(\frac{38}{-133}>\frac{-129}{344}\)

Vậy \(\frac{38}{-133}>\frac{-129}{344}\)

38/133>29/344

38/133>38/344(vì 133<344).

Mà 38/344>29/344(vì 38>29).

=>38/133>29/344.

tk mk nha các bn.

-chúc ai tk cho mk học giỏi và may mắn ,thanks các bn nhìu-

c) 

\(\frac{19}{18}=1+\frac{1}{18}\)  

\(\frac{2017}{2016}=1+\frac{1}{2016}\)   

Vì \(\frac{1}{18}>\frac{1}{2016}\)  

Vậy \(\frac{19}{18}>\frac{2017}{2016}\)    

d) 

\(\frac{133}{173}=\frac{130+3}{170+3}=\frac{13+0,3}{17+0,3}\)  

Ta có : 

\(\frac{a}{b}< \frac{a+x}{b+x}\forall a;b;x>0\)  

Vậy \(\frac{13}{17}< \frac{133}{173}\)

\(\frac{200}{201}+\frac{201}{202}>\frac{200}{201+202}+\frac{201}{201+202}=\frac{200+201}{201+202}\)

\(A=\frac{54.107-53}{53.107+54}=\frac{\left(53+1\right)107-53}{53.107+54}=\frac{53.107+107-53}{53.107+54}=\frac{53.107+54}{53.107+54}=1\)

\(B=\frac{135.269-133}{134.269+135}=\frac{\left(134+1\right)269-133}{134.269+135}=\frac{134.269+269-133}{134.269+135}=\frac{134.269+136}{134.269+135}>1\)

Vậy A<B

câu a là

200/201+201/202>200/202+201/202>1 

200+201/201+201<1

=>200/201+201/202>1>200+201/201+202

\(a)\) Ta có công thức : 

\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)

Áp dụng vào ta có : 

\(A=\frac{10^{11}-1}{10^{12}-1}< \frac{10^{11}-1+11}{10^{12}-1+11}=\frac{10^{11}+10}{10^{12}+10}=\frac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}=\frac{10^{10}+1}{10^{11}+1}=B\)

\(\Rightarrow\)\(A< B\)

Vậy \(A< B\)

Chúc bạn học tốt ~