A)\(\frac{5}{9}+\left(\frac{-2}{3}\right)^2.25\%-1\frac{1}{3}\)
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S
9 tháng 3
\(a;x:\left(9\dfrac{1}{2}-\dfrac{3}{2}\right)=\dfrac{\dfrac{2}{5}+\dfrac{4}{9}-\dfrac{5}{11}}{\dfrac{8}{5}+\dfrac{16}{9}-\dfrac{20}{11}}\\ x:8=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}\cdot8=2\)
\(b;\left|2x-\dfrac{1}{3}\right|-\left(-2^2\right)=4\left(\dfrac{1}{-2}\right)^3\\ \left|2x-\dfrac{1}{3}\right|+4=-\dfrac{1}{2}\\ \left|2x-\dfrac{1}{3}\right|=-\dfrac{1}{2}-4=-\dfrac{9}{2}\\ \Rightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=-\dfrac{9}{2}\Rightarrow x=-\dfrac{25}{12}\\2x-\dfrac{1}{3}=\dfrac{9}{2}\Rightarrow x=\dfrac{29}{12}\end{matrix}\right.\)
N
20 tháng 1 2019
\(VP=1+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4023}{2011}+\frac{4024}{2012}\)
\(=1-1+\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4023}{2011}-1\right)+\left(\frac{40024}{2012}-1\right)+2012\)
\(=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}+\frac{2012}{1}\)
\(=2012.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(\Rightarrow2012=503.x\Rightarrow x=\frac{2012}{503}=4\)
\(\frac{5}{9}+\left(\frac{-2}{3}\right)^2.25\%-1\frac{1}{3}\) =\(\frac{5}{9}+\frac{4}{9}.\frac{1}{4}-\frac{4}{3}\)=\(\frac{5}{9}+\frac{1}{9}-\frac{4}{3}\)=\(\frac{2}{3}-\frac{4}{3}\)=\(\frac{-2}{3}\)
\(\frac{5}{9}+\left(\frac{-2}{3}\right)^2.25\%-1\frac{1}{3}\)
\(=\frac{5}{9}+\frac{4}{9}.\frac{1}{4}-\frac{4}{3}\)
\(=\frac{5}{9}+\frac{1}{9}-\frac{4}{3}\)
\(=\frac{2}{3}-\frac{4}{3}=\frac{-2}{3}.\)