(-1/2)-(-3/5)+(-1/9)+1/127-(+7/18)+4/35-(-2/7)=???
hêu me!!!
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\(\left(-\dfrac{1}{2}\right)-\left(-\dfrac{3}{5}\right)+\left(-\dfrac{1}{9}\right)+\dfrac{1}{127}-\dfrac{7}{18}+\dfrac{4}{35}-\left(-\dfrac{2}{7}\right)\)
\(=\left[-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right]+\left[\dfrac{3}{5}+\dfrac{4}{35}+\dfrac{2}{7}\right]+\dfrac{1}{127}\)
\(=-\dfrac{18}{18}+\dfrac{35}{35}+\dfrac{1}{127}\)
\(=-1+1+\dfrac{1}{127}\)
\(=\dfrac{1}{127}\)
\(\left(-\frac{1}{2}\right)-\left(-\frac{3}{5}\right)+\left(-\frac{1}{9}\right)+\frac{1}{127}-\frac{7}{18}+\frac{4}{35}-\left(-\frac{2}{7}\right)=-\frac{1}{2}+\frac{3}{5}-\frac{1}{9}+\frac{1}{127}-\frac{7}{18}+\frac{4}{35}+\frac{2}{7}=\frac{1}{127}\)
\(A=\left\{\left[\left(\frac{-1}{2}+\frac{-1}{9}\right)-\frac{7}{18}\right]+\left(\frac{3}{5}+\frac{4}{35}+\frac{2}{7}\right)+\frac{1}{127}\right\}\)
\(\Rightarrow A=\left\{\left[\frac{-11}{18}-\frac{7}{18}\right]+1+\frac{1}{127}\right\}\)
\(A=\left(-1\right)+1+\frac{1}{127}\)
\(A=0+\frac{1}{127}\)
\(A=\frac{1}{127}\)
Ta có: \(\frac{-1}{2}+\frac{3}{5}+\frac{-1}{9}+\frac{1}{127}+\frac{-7}{18}+\frac{4}{35}+\frac{2}{7}\)
\(=\left(\frac{3}{5}+\frac{2}{7}+\frac{4}{35}\right)+\left(\frac{-1}{2}+\frac{-1}{9}+\frac{-7}{18}\right)+\frac{1}{127}\)
\(=\left(\frac{21}{35}+\frac{10}{35}+\frac{4}{35}\right)+\left(\frac{-9}{18}+\frac{-2}{18}+\frac{-7}{18}\right)+\frac{1}{127}\)
\(=\frac{35}{35}+\frac{-18}{18}+\frac{1}{127}\)
\(=1+\left(-1\right)+\frac{1}{127}\)
\(=\frac{1}{127}\)
(\(\frac{-1}{2}\)) -(\(\frac{-3}{5}\))+(\(\frac{-1}{9}\))+\(\frac{1}{127}\)-\(\frac{7}{18}\)+\(\frac{4}{35}\)-(\(\frac{-2}{7}\))=\(\frac{448}{1143}\)