a) Ta có: \(\frac{13}{57}=1-\frac{44}{57}\)

              \(\frac{29}{73}=1-\frac{44}{73}\)

Ta thấy:    \(\frac{44}{57}>\frac{44}{73}\)\(\Rightarrow\)\(1-\frac{44}{57}< 1-\frac{44}{73}\)

Vậy   \(\frac{13}{57}< \frac{29}{73}\)

Giải:

A=10²⁰⁰⁴+1/10²⁰⁰⁵+1

10A=10²⁰⁰⁵+10/10²⁰⁰⁵+1

10A=10²⁰⁰⁵+1+9/10²⁰⁰⁵+1

10A=1+9/10²⁰⁰⁵+1

Tương tự:

B=10²⁰⁰⁵+1/10²⁰⁰⁶+1

10B=1+9/10²⁰⁰⁶+1

Vì 9/10²⁰⁰⁵+1>9/10²⁰⁰⁶+1 nên 10A>10B

⇒A>B

Chúc bạn học tốt!

\(10A=10.\dfrac{10^{2004}+1}{10^{2005}+1}=\dfrac{10^{2005}+10}{10^{2005}+1}=1+\dfrac{9}{10^{2005}+1}\\ 10B=10.\dfrac{10^{2005}+1}{10^{2006}+1}=\dfrac{10^{2006}+10}{10^{2006}+1}=1+\dfrac{9}{10^{2006}+1}\)

vì \(\dfrac{9}{10^{2005}+1}>\dfrac{9}{10^{2006}+1}\Rightarrow10A>10B\Rightarrow A>B\)

Đề bài đâu bn?

Ta có: \(10\cdot A=\dfrac{10^{2005}+10}{10^{2005}+1}=1+\dfrac{9}{10^{2005}+1}\)

\(10B=\dfrac{10^{2006}+10}{10^{2006}+1}=1+\dfrac{9}{10^{2006}+1}\)

mà \(\dfrac{9}{10^{2005}+1}>\dfrac{9}{10^{2006}+1}\)

nên 10A>10B

hay A>B

Co j góc lệch cùng trời cuối đất cú mèo

a)=

b)>

c)=

Bài 1:

\(\dfrac{-5}{18}=\dfrac{-20}{72};\dfrac{7}{-24}=\dfrac{-21}{72}.\)

\(\dfrac{-15}{-40}=\dfrac{3}{8}=\dfrac{9}{24};\dfrac{24}{-72}=\dfrac{-1}{3}=\dfrac{-8}{24}.\)

Bài 3:

a) \(\dfrac{2}{3}h=\dfrac{8}{12}h;\dfrac{3}{4}h=\dfrac{9}{12}h.\Rightarrow\dfrac{2}{3}h< \dfrac{3}{4}h.\)

b) \(\dfrac{4}{5}km/h=\dfrac{8}{10}km/h;\dfrac{9}{10}km/h.\Rightarrow\dfrac{4}{5}km/h< \dfrac{9}{10}km/h.\)