\(\frac{-3}{27}+\frac{56}{88}\) 

\(\frac{-24}{54}+\frac{35}{-126}\)

tính tổng rồi rút gọn

\(\frac{6+a}{9-a^2}+...=\frac{6}{9-a^2}+...+\frac{a^2}{9a-a^3}\ge\frac{54}{27-a^2-b^2-c^2}+\frac{\left(a+b+c\right)^2}{9\left(a+b+c\right)-\left(a^3+b^3+c^3\right)}\)

\(\ge\frac{54}{27-2\left(a+b+c\right)+3}+\frac{9}{27-3\left(a+b+c\right)+6}=\frac{54}{24}+\frac{9}{24}=\frac{21}{8}\)

\(\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)

cho các phhan số:\(\frac{-21}{27};\frac{-14}{19};\frac{-42}{-54};\frac{35}{-45};\frac{-5}{7};\frac{-28}{36}.\) những phân số nào biểu diễn số hữu tỉ \(\frac{-7}{9}?\)

cho các biểu thức

A=\(\left(\frac{1}{8\cdot14}+\frac{1}{14\cdot20}+\frac{1}{20\cdot26}+...+\frac{1}{50\cdot56}\right)\)

\(B=\left(\frac{45}{12\cdot21}+\frac{45}{21\cdot30}+\frac{40}{24\cdot34}-\frac{40}{34\cdot44}-\frac{40}{44\cdot54}-\frac{40}{54\cdot64}\right)\)

Chứng minh rằng:\(\frac{A}{B}< \frac{1}{8}\)

tìm các tích sau: a) \(\frac{16}{15}.\frac{-5}{14}.\frac{54}{24}.\frac{56}{21}\)                                b) \(\frac{7}{3}.\frac{-5}{2}.\frac{15}{21}.\frac{4}{-5}\)

a=\(\frac{3}{54}+\frac{5}{126}+\frac{7}{294}+\frac{8}{609}+\frac{13}{1218}\)

chứng minh rằng các phân số sau đây bằng nhau :

a\(\frac{41}{88};\frac{4141}{8888};\frac{414141}{888888}\)

b\(\frac{27425-27}{x99900};\frac{27425425-27425}{99900000}\)

c\(\frac{29700-54}{x30800-56};\frac{59400-108}{61600-112}\)

\(\frac{\frac{3}{67}\left(17\frac{21}{56}-13\frac{21}{45}\right):\left(\frac{3}{5.22}+\frac{54}{44.65}+\frac{18}{65.72}\right)}{\left(29^3:100-29^3:0,47\right)}\)