\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(=\frac{1\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}{4.\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}=1\)

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}\)

=1

Ta có:

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left[\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right]}=\frac{1}{4}\)

Lại có:

\(\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}=\frac{3\left[\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right]}{4\left[\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right]}=\frac{3}{4}\)

Vậy: 

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}=\frac{1}{4}+\frac{3}{4}=1\)

A=\([\)\(\frac{2}{7}\)\(\times\)(\(\frac{1}{4}-\frac{1}{3}\))\(]\)\(\div\)\([\)(\(\frac{2}{7}\times\)(\(\frac{3}{9}-\frac{2}{5}\))\(]\)
  =(\(\frac{2}{7}\times\)\(\frac{-1}{12}\))\(\div(\)\(\frac{2}{7}\times\)\(\frac{-1}{15}\))
=\(\frac{-1}{42}\)\(\div\)\(\frac{-2}{35}\)
=\(\frac{-1}{42}\)\(\times\)\(\frac{35}{-2}\)
=\(\frac{5}{12}\)

1/2+2/3+3/4+4/5+5/6+6/7+7/8+8/9+9/10x9/10

=9/10x(1/2+2/3)+(3/4+4/5)+(5/6+6/7)+(7/8+8/9)

=9/10x(1/3+3/5+5/7+7/9)

9/10x(1/3+3/5)+(5/7+7/9)

=9/10x1/5+5/9

9/50+5/9

=10

Bn Long làm đúng rồi bn   nguyễn kim arica cứ làm theo cách đó là được .

Bn nào thấy đúng thì ủng hộ nha .

hình như đề bạn sai thì phải đúng ko biết ngỏ bạn làm

cái chỗ có 2 phân số 9/10 đấy

\(B=4\cdot\left(-\frac{1}{2}\right)^3:\left(\frac{4}{5}\right)^0\cdot\frac{1}{2}-\frac{\frac{3}{5}-\frac{3}{9}+\frac{3}{13}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{13}}\)

\(=4\cdot\frac{-1}{8}:1\cdot\frac{1}{2}-\frac{3\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{13}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{13}\right)}\)

\(=-\frac{1}{4}-\frac{3}{7}=-\frac{19}{28}\)

a= (\(\frac{2}{5}\)+\(\frac{2}{9}\)+\(\frac{2}{11}\)\(\times\)\(\frac{5}{7}\)\(+\frac{7}{9}\)\(+\frac{7}{11}\)\()\)

= 1/3 - 1/3 + 5/7 - 5/7 - 7/9 + 7/9 +9/11 - 9/11 -11/13 + 11/13 +13/15

= 0 + 0 - 0 + 0 -0 + 13/15

= 0 + 13/15

= 13/15

\(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}+\frac{13}{15}+\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+\frac{3}{5}-\frac{1}{3}\)

\(=\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{3}{5}-\frac{3}{5}\right)+\left(\frac{5}{7}-\frac{5}{7}\right)+\left(\frac{7}{9}-\frac{7}{9}\right)+\left(\frac{9}{11}-\frac{9}{11}\right)+\left(\frac{11}{13}-\frac{11}{13}\right)+\frac{13}{15}\)

\(=0+0+0+0+0+0+\frac{13}{15}\)

\(=\frac{13}{15}\)

\(N=\frac{1}{6}.\frac{5}{9}+\frac{5}{9}.\frac{2}{7}+\frac{1}{9}.\frac{5}{7}-\frac{5}{9}.\frac{3}{7}\)

\(N=\frac{1}{6}.\left(\frac{5}{9}+\frac{1}{9}\right).\left(\frac{2}{7}+\frac{5}{7}-\frac{3}{7}\right)\)

\(N=\frac{1}{6}.\frac{6}{9}.\frac{4}{7}\)

\(N=\frac{1}{9}.\frac{4}{7}\)

\(N=\frac{4}{63}\)

\(N=\frac{1}{6}.\frac{5}{9}+\frac{5}{9}.\frac{2}{7}+\frac{1}{9}.\frac{5}{7}-\frac{5}{9}.\frac{3}{7}\)

\(N=\frac{1}{6}.\frac{5}{9}+\frac{5}{9}.\frac{2}{7}+\frac{5}{9}.\frac{1}{7}-\frac{5}{9}.\frac{3}{7}\)

\(N=\frac{5}{9}.\left(\frac{1}{6}+\frac{2}{7}+\frac{1}{7}-\frac{3}{7}\right)\)

\(N=\frac{5}{9}.\frac{1}{6}\)

\(N=\frac{5}{54}\)