\(\frac{52}{75}=\frac{52.101}{75.101}=\frac{5252}{7575};\frac{52}{75}=\frac{52.10101}{75.10101}=\frac{525252}{757575}\)

\(\frac{13}{15}=\frac{13.101}{15.101}=\frac{1313}{1515};\frac{13}{15}=\frac{13.10101}{15.10101}=\frac{131313}{151515}\)

\(\frac{ab}{cd}=\frac{101ab}{101cd}=\frac{abab}{cdcd};\frac{ab}{cd}=\frac{10101ab}{10101cd}=\frac{ababab}{cdcdcd}\)

ai k minh minh k lai

chia các vế với 1001 và 100001

Ta có abab/cdcd=abab:101/cdcd:101=ab/cd

          ababab/cdcdcd=ababab:10101/cdcdcd:10101=ab/cd

Vì ab/cd=ab/cd nên  abab/cdcd= ababab/cdcdcd

\(\frac{abab}{cdcd}=\frac{ab.101}{cd.101}=\frac{ab}{cd}\)

\(\frac{ababab}{cdcdcd}=\frac{ab.10101}{cd.10101}=\frac{ab}{cd}\)

VẬY \(\frac{abab}{cdcd}=\frac{ababab}{cdcdcd}\)

A = 1/2^2 + 1/3^2 + 1/4^2 + ... + 1/100^2

1/2^2 < 1/1*2

1/3^2 < 1/2*3

1/4^2 < 1/3*4

...

1/100^2 < 1/99*100

=> A < 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/99*100

=> A < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100

=> A < 1 - 1/100

=> A < 1

minh deo can ban k dau :((

\(a,\frac{1}{2}x+\frac{3}{5}(x-2)=3\)

\(\Rightarrow\frac{1}{2}x+\frac{3}{5}x-\frac{6}{5}=3\)

\(\Rightarrow\left[\frac{1}{2}+\frac{3}{5}\right]x=3+\frac{6}{5}\)

\(\Rightarrow\left[\frac{5}{10}+\frac{6}{10}\right]x=\frac{21}{5}\)

\(\Rightarrow\frac{11}{10}x=\frac{21}{5}\)

\(\Rightarrow x=\frac{21}{5}:\frac{11}{10}=\frac{21}{5}\cdot\frac{10}{11}=\frac{21}{1}\cdot\frac{2}{11}=\frac{42}{11}\)

Vậy x = 42/11

1) 

a) \(\frac{9^{14}.25^5.8^7}{18^{12}.625^3.24^3}=\frac{3^{28}.5^{10}.2^{21}}{2^{21}.3^{24}.5^{12}.3^3.2^9}=\frac{3}{5^2}=\frac{3}{25}\)

Bài 2:

\(\frac{abab}{cdcd}=\frac{ab.101}{cd.101}=\frac{ab}{cd};\frac{ababab}{cdcdcd}=\frac{ab.10101}{cd.10101}=\frac{ab}{cd}\)

Vậy \(\frac{ab}{cd}=\frac{abab}{cdcd}=\frac{ababab}{cdcdcd}\)

Ta co :

\(\frac{abab}{cdcd}\) va \(\frac{ababab}{cdcdcd}\)

\(\Rightarrow\frac{abab}{cdcd}=\frac{ab}{cd}\)

\(\Rightarrow\frac{ababab}{cdcdcd}=\frac{ab}{cd}\)

Ta thay :\(\frac{ab}{cd}=\frac{ab}{cd}\)

Vay :\(\frac{abab}{cdcd}=\frac{ababab}{cdcdcd}\)

cấm ai đc copy bài tớ

Ta co :

$\genfrac{}{}{0ex}{}{abab}{cdcd}$ va $\genfrac{}{}{0ex}{}{ababab}{cdcdcd}$

$\Rightarrow \genfrac{}{}{0ex}{}{abab}{cdcd}=\genfrac{}{}{0ex}{}{ab}{cd}$

$\Rightarrow \genfrac{}{}{0ex}{}{ababab}{cdcdcd}=\genfrac{}{}{0ex}{}{ab}{cd}$

Ta thay :$\genfrac{}{}{0ex}{}{ab}{cd}=\genfrac{}{}{0ex}{}{ab}{cd}$

Vay :$\genfrac{}{}{0ex}{}{abab}{cdcd}=\genfrac{}{}{0ex}{}{ababab}{cdcdcd}$
 

có bn ạ

mk nghĩ zậy

có bạn ạ

mkk o chắc nữa

Hai phân số sau có bằng nhau không?

$\frac{\overline{abab}}{\overline{cdcd}}=\frac{\overline{ababab}}{\overline{cdcdcd}}$=\(\frac{ab}{cd}\)

\(\Rightarrow\) Hai phan so đều bằng nhau.

cảm ơn nhé tớ nghĩ ra rồi!

a)  a b = a . ( − 1 ) b . ( − 1 ) = − a − b

b) ta có:

a b a b ¯ c d c d ¯ = a b a b ¯ : 101 c d c d ¯ : 101 = a b ¯ c d ¯ ; a b a b a b ¯ c d c d ¯ c d = a b a b a b ¯ : 10101 c d c d ¯ c d:10101 = a b ¯ c d ¯

do đó:  a b a b ¯ c d c d ¯ = a b a b a b ¯ c d c d ¯ c d

c)  a b a b ¯ a b a b a b ¯ = a b a b ¯ : a b ¯ a b a b a b ¯ : a b ¯ = 101 10101