Tính
a) \(\frac{2}{3}+\frac{1}{3}.\left(\frac{-4}{9}+\frac{5}{6}\right):\frac{7}{12}\)
b) \(\frac{298}{719}:\left(\frac{1}{4}+\frac{1}{12}-\frac{1}{3}\right)-\frac{2011}{2012}\)
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\(\frac{298}{719}:\left(\frac{1}{4}+\frac{1}{12}-\frac{1}{3}\right)-\frac{2011}{2012}\)
\(=\frac{298}{719}:0-\frac{2011}{2012}\)
Giá trị phép tính không tồn tại
\(\frac{298}{719}\left(\frac{1}{4}+\frac{1}{12}-\frac{1}{3}\right)-\frac{2011}{2012}\)
\(=\frac{298}{719}\cdot0-\frac{2011}{2012}\)
\(=0-\frac{2011}{2012}=-\frac{2011}{2012}\)
a)
\(\begin{array}{l}\frac{1}{9} - 0,3.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{10}}.\frac{5}{9} + \frac{1}{3}\\ = \frac{1}{9} - \frac{3}{{2.5}}.\frac{5}{{3.3}} + \frac{1}{3}\\ = \frac{1}{9} - \frac{1}{6} + \frac{1}{3}\\ = \frac{2}{{18}} - \frac{3}{{18}} + \frac{6}{{18}}\\ = \frac{5}{{18}}\end{array}\)
b)
\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^2} + \frac{1}{6} - {\left( { - 0,5} \right)^3}\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{2}} \right)^3\\ = \frac{4}{9} + \frac{1}{6} - \left( {\frac{{ - 1}}{8}} \right)\\ = \frac{4}{9} + \frac{1}{6} + \frac{1}{8}\\ = \frac{{32}}{{72}} + \frac{{12}}{{72}} + \frac{9}{{72}}\\ = \frac{{53}}{{72}}\end{array}\)
\(\frac{27.\left(18+103-120\right)}{33.\left(15+12\right)}\)=\(\frac{27.1}{33.27}\)=\(\frac{1}{33}\)
mik dang ban moi giai duoc mot bai ha, sorry
\(VP=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)
\(=1-1+\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4023}{2011}-1\right)+\left(\frac{40024}{2012}-1\right)+2012\)
\(=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}+\frac{2012}{1}\)
\(=2012.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(\Rightarrow2012=503.x\Rightarrow x=\frac{2012}{503}=4\)
\(\frac{2}{3}+\frac{1}{3}\left(\frac{-4}{9}+\frac{5}{6}\right)\div\frac{7}{12}\)
\(=\frac{2}{3}+\frac{1}{3}\left(\frac{-4}{9}+\frac{5}{6}\right)\times\frac{12}{7}\)
\(=\frac{2}{3}+\frac{4}{7}\left(\frac{-4}{9}+\frac{5}{6}\right)\)
\(=\frac{2}{3}+\frac{4}{7}\left(\frac{-8}{18}+\frac{15}{18}\right)\)
\(=\frac{2}{3}+\frac{4}{7}.\frac{-7}{18}\)
\(=\frac{6}{9}+\frac{-2}{9}\)
\(=\frac{4}{9}\)
\(\frac{298}{719}\div\left(\frac{1}{4}+\frac{1}{12}-\frac{1}{3}\right)-\frac{2011}{2012}\)
\(=\frac{298}{719}\div\left(\frac{3}{12}+\frac{1}{12}-\frac{4}{12}\right)-\frac{2011}{2012}\)
\(=\frac{298}{719}\div0-\frac{2011}{2012}\)
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