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b)\(\left(2016.1017+2017.2018\right).\left(1+\frac{1}{2}:\frac{3}{2}-\frac{4}{3}\right)\)
\(\left(2016.2017+2017.2018\right)\left(1+\frac{1}{3}-\frac{4}{3}\right)\)
\(\left(2016.2017+2017.2018\right).\left(\frac{4}{3}-\frac{4}{3}\right)\)
\(\left(2016.2017+2017.2018\right).0\)
\(=0\)
\(\frac{18}{11}+\left(\frac{7}{4}-\frac{3}{5}\right):\frac{1}{2}\)
\(\frac{18}{11}+\frac{23}{20}:\frac{1}{2}\)
\(\frac{18}{11}+\frac{23}{20}\times\frac{2}{1}\)
\(\frac{18}{11}+\frac{23}{10}\)
\(\frac{433}{110}\)
rồi, h thì dễ hỉu @@
\(\left(\frac{74}{7}+\frac{19}{5}\right)-\left(\frac{39}{7}-\frac{6}{5}\right)=\frac{74}{7}+\frac{19}{5}-\frac{39}{7}+\frac{6}{5}=\left(\frac{74}{7}-\frac{39}{7}\right)+\left(\frac{19}{5}+\frac{6}{5}\right)=\frac{35}{7}+\frac{25}{5}=5+5=10\)
Ta có \(\left(x-\frac{1}{2}\right).\frac{5}{2}+\frac{1}{2}=\frac{7}{4}\)
\(\Rightarrow\left(x-\frac{1}{2}\right).\frac{5}{2}=\frac{7}{4}-\frac{1}{2}\)
\(\Rightarrow\left(x-\frac{1}{2}\right).\frac{5}{2}=\frac{5}{4}\)
\(\Rightarrow x-\frac{1}{2}=\frac{5}{4}\div\frac{5}{2}\)
\(\Rightarrow x-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow x=1\)
(x-\(\frac{1}{2}\)).\(\frac{5}{2}\)+\(\frac{1}{2}\)=\(\frac{7}{4}\)
(x-\(\frac{1}{2}\)).\(\frac{5}{2}\)=\(\frac{7}{4}\)- \(\frac{1}{2}\)
(x-\(\frac{1}{2}\)).\(\frac{5}{2}\)= \(\frac{7}{4}\)- \(\frac{2}{4}\)
(x-\(\frac{1}{2}\)).\(\frac{5}{2}\)\(\frac{5}{2}\)= \(\frac{5}{4}\)
(x-\(\frac{1}{2}\)) = \(\frac{5}{4}\): \(\frac{5}{2}\)= \(\frac{5}{4}\). \(\frac{2}{5}\)= \(\frac{1}{2}\)
x = \(\frac{1}{2}\)+ \(\frac{1}{2}\)=\(\frac{1}{4}\)
vậy x\(\frac{7}{4}\)\(\frac{7}{4}\)
\(=\frac{4}{7}+\frac{4}{7}-\frac{15}{28}\)
=\(\frac{4}{7}.\left(1+1-0,9375\right)\)
=\(\frac{17}{28}\approx0,61\)