\(\left(\frac{-1}{5}+\frac{3}{12}\right)+\frac{-3}{4}\)
giúp mình nha . Cảm ơn !
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\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)
\(\frac{-4}{11}.\frac{2}{5}+\frac{6}{11}.\frac{-3}{10}\)
=\(\frac{2}{11}.\left(-2\right).\frac{2}{5}+\frac{2}{11}.3.\frac{-3}{10}\)
=\(\frac{2}{11}.\frac{-4}{5}+\frac{2}{11}.\frac{-9}{10}\)
=\(\frac{2}{11}.\left(\frac{-4}{5}+\frac{-9}{10}\right)\)
=\(\frac{2}{11}.\frac{-17}{10}\)
=\(\frac{-17}{55}\)
\(\left(\frac{2}{3}-1\frac{1}{2}\right):\frac{4}{3}+\frac{1}{2}\)
=\(\left(\frac{2}{3}-\frac{3}{2}\right)\times\frac{3}{4}+\frac{1}{2}\)
=\(\frac{-5}{6}\times\frac{3}{4}+\frac{1}{2}\)
=\(\frac{-5}{8}+\frac{4}{8}\)
=\(\frac{-1}{8}\)
Ai thấy đúng thì *******
\(\left(\frac{2}{3}-1\frac{1}{2}\right):\frac{4}{3}+\frac{1}{2}\)
\(=\left(\frac{2}{3}-\frac{3}{2}\right):\frac{4}{3}+\frac{1}{2}\)
\(=\left(\frac{4}{6}-\frac{9}{6}\right):\frac{4}{3}+\frac{1}{2}\)
\(=\frac{-5}{6}:\frac{4}{3}+\frac{1}{2}\)
\(=\frac{-5}{6}.\frac{3}{4}+\frac{1}{2}\)
\(=\frac{-5}{8}+\frac{1}{2}\)
\(=\frac{-5}{8}+\frac{4}{8}\)
\(=\frac{1}{8}\)
\(\left(\frac{-1}{5}+\frac{3}{12}\right)+\frac{-3}{4}\)
\(=\left(\frac{-1}{5}+\frac{1}{4}\right)+\frac{-3}{4}\)
\(=\frac{-1}{5}+\frac{1}{4}+\frac{-3}{4}\)
\(=\frac{-1}{5}+\frac{-1}{2}\)
\(=\frac{-2}{10}+\frac{-5}{10}\)
\(=\frac{-7}{10}\)
de bam may la ra