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34 −35 +37 +311 134 −135 +137 +1311
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{11}\right)}{13.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{11}\right)}\)
\(=\frac{3}{13}\)
â, -4/9(7/15+8/15)=-4/9
b,-5/4(16/25+9/25)=-5/4
,.....
dài quá mik làm ko hết
hok tốt
\(a.\frac{108}{119}.\frac{107}{211}+\frac{108}{119}.\frac{104}{211}=\frac{108}{119}.\left(\frac{107}{211}+\frac{104}{211}\right)=\frac{108}{119}.1=108\)
Tính tất cả ra thì được:
\(=\frac{\frac{973}{60}}{\frac{139}{60}}:\frac{\frac{1255}{221}}{\frac{1506}{221}}+\frac{5858}{5050}\)
\(=\frac{\frac{139}{60}}{\frac{973}{60}}.\frac{\frac{1506}{221}}{\frac{1255}{221}}+\frac{5858}{5050}\)
Tính tử và mẫu dần rồi ra ( phần này dễ mà )
Ta được: ( mình chỉ lấy 2 chữ số phần thập phân thôi )
\(=\frac{1578}{9209}+\frac{5858}{5050}\)
= 133/100
End
2: \(=\dfrac{0.8}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\dfrac{71}{75}\cdot\dfrac{7}{4}}{\dfrac{119}{36}\cdot\dfrac{36}{17}}\)
\(=\dfrac{4}{5}\cdot\dfrac{5}{3}+\dfrac{71}{300}=\dfrac{471}{300}=\dfrac{157}{100}\)
3: \(=\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\)
=2/7-2/7=0
\(a,\frac{7}{12}\cdot\frac{6}{11}+\frac{7}{12}\cdot\frac{5}{11}+2\frac{7}{12}\)
\(=\frac{7}{12}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+2\frac{7}{12}\)
\(=\frac{7}{12}+\frac{31}{12}\)
\(=\frac{38}{12}=\frac{19}{6}\)
\(b,\frac{-5}{9}\cdot\frac{-6}{13}+\frac{5}{-9}\cdot\frac{-5}{13}-\frac{5}{9}\)
\(=\frac{-5}{9}\cdot\frac{-6}{13}+\frac{-5}{9}\cdot\frac{-5}{13}+\frac{-5}{9}\cdot1\)
\(=\frac{-5}{9}\cdot\left(\frac{-6}{13}+\frac{-5}{13}+1\right)\)
\(=\frac{-5}{9}\cdot\left(\frac{-11}{13}+1\right)\)
\(=\frac{-5}{9}\cdot\frac{2}{13}\)
\(=\frac{-10}{117}\)
\(c,\)\(0,8\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-1\frac{2}{5}\)
\(=\frac{4}{5}\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-\frac{7}{5}\)
\(=\frac{4}{5}\cdot\left(\frac{-15}{14}-\frac{13}{14}\right)-\frac{7}{5}\)
\(=\frac{4}{5}\cdot\left(-2\right)-\frac{7}{5}\)
\(=\frac{-8}{5}-\frac{7}{5}\)
\(=-3\)
\(d,\)\(75\%\cdot\frac{6}{7}+5\%\cdot\frac{6}{7}+\frac{7}{10}\cdot1\frac{1}{7}\)
\(=\frac{3}{4}\cdot\frac{6}{7}+\frac{1}{20}\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)
\(=\left(\frac{3}{4}+\frac{1}{20}\right)\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)
\(=\frac{4}{5}\cdot\frac{6}{7}+\frac{4}{5}\cdot1\)
\(=\frac{4}{5}\cdot\left(\frac{6}{7}+1\right)\)
\(=\frac{4}{5}\cdot\frac{13}{7}\)
\(=\frac{52}{35}\)
Đặt \(A=\frac{15+\frac{15}{7}-\frac{15}{11}+\frac{15}{2009}-\frac{15}{13}}{\frac{4}{2009}-\frac{4}{13}+\frac{4}{7}-\frac{4}{11}+4}\)
\(=\frac{15\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{2009}-\frac{1}{13}\right)}{4\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{2009}-\frac{1}{13}\right)}\)
\(=\frac{15}{4}\)
Đặt \(B=\frac{5\cdot2010-1996}{14+4\cdot2010}\)
\(=\frac{5\left(1996+4\right)-1996}{14+4\cdot2010}\)
\(=\frac{5\cdot1996+20-1996}{14+4\left(1996+4\right)}\)
\(=\frac{4\cdot1996+20}{4\cdot1996+30}\)
\(\Rightarrow A\cdot B=\frac{4\cdot1996+20}{4\cdot1996+30}\cdot\frac{15}{4}=\frac{15\cdot4\left(1996+5\right)}{4\left(4\cdot1996+30\right)}=\frac{15\left(1996+5\right)}{4\cdot1996+30}=\frac{30015}{8004}\)
mặc dầu ko khoa học lắm nhưng mình thấy cũng được đấy