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\(\dfrac{-4}{12}+\dfrac{18}{45}+\dfrac{-6}{9}+\dfrac{-21}{35}+\dfrac{6}{30}\)
\(=\dfrac{-1}{3}+\dfrac{2}{5}+\dfrac{-2}{3}+\dfrac{-3}{5}+\dfrac{1}{5}\)
\(=\left(\dfrac{-1}{3}+\dfrac{-2}{3}\right)+\left(\dfrac{2}{5}+\dfrac{-3}{5}+\dfrac{1}{5}\right)\)
\(=-1+0\)
\(=-1\)
\(\#PeaGea\)
\(-\frac{4}{12}+\frac{18}{45}-\frac{6}{9}-\frac{21}{35}+\frac{6}{30}\)
\(=\frac{-1}{3}+\frac{2}{5}-\frac{1}{3}-\frac{3}{5}+\frac{1}{5}\)
\(=0\)
\(E=\dfrac{-1}{3}-\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{1}{5}=-1\)
Ta có: \(\dfrac{-4}{12}+\dfrac{18}{45}+\dfrac{-6}{9}+\dfrac{21}{35}+\dfrac{6}{30}\)
\(=\dfrac{-1}{3}+\dfrac{-2}{3}+\dfrac{2}{5}+\dfrac{3}{5}+\dfrac{6}{30}\)
\(=-1+1+\dfrac{1}{5}\)
\(=\dfrac{1}{5}\)
\(A=\frac{-5}{7}+\frac{3}{4}+\frac{-1}{5}+\frac{-2}{7}+\frac{1}{4}\)
\(A=\left(\frac{-5}{7}+\frac{-2}{7}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)+\frac{-1}{5}\)
\(A=-1+1+\frac{-1}{5}\)
\(A=\frac{-1}{5}\)
\(B=\frac{-4}{12}+\frac{18}{45}+\frac{-6}{9}+\frac{-21}{35}+\frac{6}{30}\)
\(B=\frac{-1}{3}+\frac{2}{5}+\frac{-2}{3}+\frac{-3}{5}+\frac{1}{5}\)
\(B=\left(\frac{-1}{3}+\frac{-2}{3}\right)+\left(\frac{2}{5}+\frac{-3}{5}+\frac{1}{5}\right)\)
\(B=-1+0\)
\(B=-1\)
FF = \(\frac{-4}{12}+\frac{18}{45}+\frac{-6}{9}+\frac{-21}{35}+\frac{6}{30}\)
F = \(\frac{-1}{3}+\frac{6}{15}+\frac{-2}{3}+\frac{-3}{5}+\frac{3}{15}\)
F = \((\frac{-1}{3}+\frac{-2}{3})+(\frac{6}{15}+\frac{3}{15})+\frac{-3}{5}\)
F = \(\frac{-3}{3}+\frac{9}{15}+\frac{-3}{5}\)
F = \(-1+\frac{3}{5}+\frac{-3}{5}\)
F =\(-1+0\)
F = \(-1\)
Vậy F = \(-1\)