\(=\frac{29}{5}\)

chúc bạn hok tốt nhé

\(\frac{18}{15}-\frac{85}{100}+\frac{42}{10}+\frac{18}{9}-\frac{12}{16}\)

= \(\frac{7}{20}+\frac{42}{10}+\frac{18}{9}-\frac{12}{16}\)

= \(\frac{23}{20}+\frac{18}{9}-\frac{12}{16}\)

= \(\frac{63}{20}-\frac{12}{16}\)

= \(\frac{12}{5}\).

91/20

A=15x(1/7-1/12-1/98)/18(1/7-1/12-1/98)

A=15/18

A=5/6

h nhe!!!

\(A=\frac{15\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}{18\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}\)

\(A=\frac{15}{18}=\frac{5}{6}\)

\(A=\frac{15-\frac{15}{7}-\frac{15}{12}-\frac{15}{98}}{18-\frac{18}{7}-\frac{18}{12}-\frac{18}{98}}\)

\(\Rightarrow A=\frac{15\left(\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}{18\left(\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}\)

\(\Rightarrow A=\frac{15}{18}=\frac{5}{6}\)

mk làm chuẩn rồi nha

\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}\)

Tách tổng trên thành 2 nhóm, ta được :

\(\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)\)

Mà \(\frac{1}{41}>\frac{1}{42}>...>\frac{1}{60}\); \(\frac{1}{61}>\frac{1}{62}>...>\frac{1}{80}\)

\(\Rightarrow\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\right)\)

\(>\frac{1}{60}.20+\frac{1}{80}.20=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)( đpcm )

\(A=\frac{15\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}{18\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{18}\right)}=\frac{15}{18}\)

\(A=\frac{15\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}{18\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}=\frac{15}{18}=\frac{5}{6}\)

Ta có:

 A = [15 x (1-1/7-1/12-1/98)] / [ 18 x (1-1/7-1/12-1/98)]

  = 15/18 = 5/6

\(A=\frac{15\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}{18\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}=\frac{15}{18}=\frac{15:3}{18:3}=\frac{5}{6}\)

k cho mk nha

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

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

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

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

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

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

=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: 

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

\(\frac{12}{18}+\frac{1}{3}+\frac{1}{7}+\frac{2}{8}+\frac{27}{36}+\frac{42}{49}\)

\(=\frac{2}{3}+\frac{1}{3}+\frac{1}{7}+\frac{1}{4}+\frac{3}{4}+\frac{6}{7}\)

\(=\left(\frac{2}{3}+\frac{1}{3}\right)+\left(\frac{1}{7}+\frac{6}{7}\right)+\left(\frac{1}{4}+\frac{3}{4}\right)\)

\(=1+1+1\)

\(=3\)

\(a,=\left(\frac{15}{12}-\frac{3}{12}\right)+\left(\frac{5}{13}-\frac{18}{13}\right)\)

\(=1+-1\)

\(=0\)