\(\frac{4}{7}\)+ \(\frac{14}{4}\)=???
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\(\frac{1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}}{4+\frac{4}{7}+\frac{4}{7^2}-\frac{4}{7^3}}.\frac{858585}{313131}.\left(-1\frac{14}{17}\right)\)\
\(=\frac{1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}}{4.\left(1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}\right)}.\frac{85}{31}.\left(-\frac{31}{17}\right)\)
\(=\frac{1}{4}.\frac{85}{31}.\left(-\frac{31}{17}\right)\)
\(=-\frac{5}{4}\)
\(=\frac{1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}}{4\left(1+\frac{1}{7}+\frac{1}{7^2}-\frac{1}{7^3}\right)}.\frac{85.10101}{31.10101}.\left(-\frac{31}{17}\right).\)
\(=\frac{1}{4}.\frac{3.17}{31}.\left(-\frac{31}{17}\right)=-\frac{3}{4}\)
\(A=\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)
\(A=\frac{-4}{5}+\frac{4}{3}+\frac{-5}{4}+\frac{14}{5}-\frac{7}{3}\)
\(A=\left(\frac{-4}{5}+\frac{14}{5}\right)+\left(\frac{4}{3}-\frac{7}{3}\right)+\frac{-5}{4}\)
\(A=2+\left(-1\right)+\frac{-5}{4}\)
\(A=\frac{-1}{4}\)
A=\(\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)\(\frac{7}{3}\)
=\(\frac{-4}{5}+\frac{4}{3}+\frac{-5}{4}+\frac{14}{5}+\frac{-7}{3}\)=\(\left(\frac{-4}{5}+\frac{14}{5}\right)+\left(\frac{4}{3}+\frac{-7}{3}\right)+\frac{-5}{4}\)
=\(\frac{10}{5}+\frac{-3}{3}+\frac{-5}{4}\)=\(2-1+\frac{-5}{4}\)=\(1+\frac{-5}{4}\)=\(\frac{4}{4}+\frac{-5}{4}\)=\(\frac{4-5}{4}\)=\(\frac{-1}{4}\)
\(\frac{14}{6}+\frac{1}{3}+\frac{14}{14}+\frac{14}{9}+\frac{7}{15}+\frac{6}{4}\)
\(=\frac{14}{6}+\frac{1}{3}+\frac{14}{9}+\frac{7}{15}+\frac{14}{14}+\frac{6}{4}\)
\(=\frac{42+6+28}{18}+\frac{7}{15}+\frac{2}{2}+\frac{3}{2}\)
\(=\frac{76}{18}+\frac{7}{15}+\frac{5}{2}\)
\(=\frac{76}{18}+\frac{5}{2}+\frac{7}{15}\)
\(=\frac{76+45}{18}+\frac{7}{15}\)
\(=\frac{121}{18}+\frac{7}{15}\)
\(=\frac{605+42}{90}\)
\(=\frac{647}{90}\)
\(\frac{14}{6}+\frac{1}{3}+\frac{14}{14}+\frac{14}{9}+\frac{7}{15}+\frac{6}{4}\)
\(=(\frac{7}{3}+\frac{1}{3})+1+(\frac{14}{9}+\frac{7}{15})+\frac{3}{2}\)
\(=\frac{8}{3}+1+\frac{91}{45}+\frac{3}{2}\)
\(=2+\frac{2}{3}+1+2+\frac{1}{45}+1+\frac{1}{2}\)
\(=\left(\frac{2}{3}+\frac{1}{45}+\frac{1}{2}\right)+\left(2+1+2+1\right)\)
\(=1+\frac{17}{90}+6\)
\(=7+\frac{17}{90}\)
\(=7\frac{17}{90}\)
Mình làm như thế này nek
\(\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{0,75+\frac{9}{7}-2\frac{2}{5}}+\frac{\frac{3}{14}-\frac{2}{10}+\frac{5}{18}+\frac{7}{66}}{\frac{6}{7}-\frac{4}{5}+\frac{10}{9}+\frac{14}{33}}\)
\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{\frac{2}{4}+\frac{9}{7}-\frac{12}{5}}+\frac{\frac{1}{2}\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}{2\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}\)
\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{3\cdot\left(\frac{1}{4}+\frac{3}{7}-\frac{4}{5}\right)}+\frac{\frac{1}{2}}{2}\)
\(=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
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
\(\frac{9}{14}.y=\frac{7.4}{14.7}\)
\(\frac{9}{14}.y=\frac{1.2}{7.1}\)
\(\frac{9}{14}.y=\frac{2}{7}\)
\(y=\frac{2}{7}:\frac{9}{14}=\frac{2}{7}.\frac{14}{9}\)
\(y=\frac{2.2}{1.9}=\frac{4}{9}\)
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Ta có : \(\frac{9}{14}y=\frac{7}{14}.\frac{4}{7}\)
\(\Rightarrow\frac{9}{14}y=\frac{2}{7}\)
\(\Rightarrow y=\frac{2}{7}.\frac{14}{9}\)
\(\Rightarrow y=\frac{4}{9}\)
A= (-4/5+4/3)+(-5/4+14/5)-7/3
= 8/15+31/20-7/3
= 25/12-7/3
= -1/4
B= 8/3.2/5.3/8.10.19/92
= 16/15.3/8.10.19/92
= 2/5.10.19/92
= 4.19/92
= 19/23
C= \(\frac{-5}{7}\).\(\frac{2}{11}\)+\(\frac{-5}{7}\).\(\frac{9}{14}\)+1\(\frac{5}{7}\)
=\(\frac{-5}{7}\).\(\frac{2}{11}\)+\(\frac{-5}{7}\).\(\frac{9}{14}\)+\(\frac{12}{7}\)
= \(\frac{-10}{77}\)+\(\frac{-5}{7}\).\(\frac{9}{14}\)+\(\frac{12}{7}\)
= \(\frac{-10}{77}\)+\(\frac{-45}{98}\)+\(\frac{12}{7}\)
= \(\frac{-635}{1078}\)+\(\frac{12}{7}\)
= \(\frac{1213}{1078}\)
\(\frac{4}{7}+\frac{14}{4}=\frac{16}{28}+\frac{98}{28}=\frac{16+98}{28}=\frac{114}{28}=\frac{57}{14}\)
Chúc bạn học tốt !~!
\(\frac{4}{7}+\frac{14}{4}=\frac{16}{28}+\frac{98}{28}=\frac{114}{28}=\frac{57}{14}\)