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\(=-\frac{3}{4}+\frac{6}{5}-\left(\frac{-1}{8}\right)\))
=\(-\frac{3}{4}+\frac{6}{5}+\frac{1}{8}\)
=\(\frac{9}{20}+\frac{1}{8}\)
=\(\frac{23}{40}\)
Vậy..........
\(\frac{75}{100}-1\frac{1}{2}+0,5:\frac{5}{12}-\left(-\frac{1}{2}\right)^3\)
= \(\frac{3}{4}-\frac{3}{2}+\frac{6}{5}+\frac{1}{8}\)
= \(\frac{30}{40}-\frac{60}{40}+\frac{48}{40}+\frac{5}{40}\)
= \(\frac{23}{40}\)
\(\frac{5}{7}\times\frac{1}{3}-\frac{5}{7}\times\frac{1}{4}-\frac{5}{7}\times\frac{1}{2}\)
\(=\frac{5}{7}\times\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{2}\right)\)
\(=\frac{5}{7}\times\left(\frac{4}{12}-\frac{3}{12}-\frac{6}{12}\right)\)
\(=\frac{5}{7}\times\left(\frac{4-3-6}{12}\right)\)
\(=\frac{5}{7}\times\frac{-5}{12}\)
\(=\frac{5\times\left(-5\right)}{7\times12}\)
\(=\frac{-25}{84}\)
\(\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{2}\)
= \(\frac{5}{7}.\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{2}\right).1\)
\(=\frac{5}{7}.\frac{-5}{12}\)
\(=-\frac{25}{84}\)
\(C=(\frac{2}{3}-\frac{1}{4}+\frac{5}{11}):(\frac{5}{12}+1-\frac{7}{11})\)
\(=\left(\frac{88}{132}-\frac{33}{132}+\frac{60}{132}\right):\left(\frac{55}{132}+\frac{132}{132}-\frac{84}{132}\right)=\left(\frac{115}{132}\right):\frac{103}{132}=\frac{115}{132}.\frac{132}{103}=\frac{115}{103}\)
\(D=1\frac{1}{3}+\frac{1}{8}:\left(0,75-\frac{1}{2}\right)-\frac{25}{100}.\frac{1}{2}=\frac{1}{3}+\frac{1}{8}:\frac{1}{4}-\frac{1}{8}=\frac{1}{3}+\frac{1}{2}-\frac{1}{8}=\frac{8+12-3}{24}=\frac{17}{24}\)
\(E=\left(-\frac{1}{2}\right)^2-\left(-2\right)^2-5^0=\frac{1}{4}-4-1=\frac{1-16-4}{4}=\frac{-19}{4}\)
\(a,\frac{-7}{25}.\frac{11}{13}+\frac{-7}{25}.\frac{2}{13}-\frac{18}{25}\)
\(=\frac{-7}{25}.\left(\frac{11}{13}+\frac{2}{13}\right)-\frac{18}{25}=\frac{-7}{25}-\frac{18}{25}=-1\)
\(b,\frac{5}{7}.\frac{1}{3}-\frac{5}{7}.\frac{1}{4}-\frac{5}{7}.\frac{1}{12}=\frac{5}{7}.\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)=\frac{5}{7}.\left(\frac{4}{12}-\frac{3}{12}-\frac{1}{12}\right)\)
\(=\frac{5}{7}.0=0\)
c)\(5\frac{2}{5}.4\frac{2}{7}+5\frac{5}{7}.5\frac{2}{5}=\frac{27}{5}.\frac{30}{7}+\frac{40}{7}.\frac{27}{5}=\frac{27}{5}.\left(\frac{30}{7}+\frac{40}{7}\right)\)
\(=\frac{27}{5}.10=27.2=54\)
\(d,75\%-1\frac{1}{2}+0,5:\frac{5}{12}-\left(\frac{-1}{2}\right)^2=\frac{3}{4}-\frac{3}{2}+\frac{1}{2}.\frac{12}{5}-\frac{1}{4}\)
\(=\left(\frac{3}{4}-\frac{1}{4}\right)-\frac{3}{2}+\frac{6}{5}=\frac{1}{2}-\frac{3}{2}+\frac{6}{5}=-1+\frac{6}{5}=\frac{-5}{5}+\frac{6}{5}=\frac{1}{5}\)
\(=\frac{75}{100}-\frac{3}{2}+\frac{1}{2}:\frac{5}{12}-\left(-\frac{1}{3}\right)^2\)
\(=\frac{3}{4}-\frac{3}{2}+\frac{6}{5}-\frac{1}{9}\)
\(=\frac{-3}{4}+\frac{6}{5}-\frac{1}{9}\)
\(=\frac{9}{20}-\frac{1}{9}\)
\(=\frac{61}{180}\)
\(\frac{75}{100}-1\frac{1}{2}+0,5:\frac{5}{12}-\left(-\frac{1}{2}\right)^3\)
= \(\frac{3}{4}-\frac{3}{2}+\frac{1}{2}:\frac{5}{12}-\frac{-1}{8}\)
= \(\frac{3}{4}-\frac{6}{4}+\frac{1}{2}\cdot\frac{12}{5}-\frac{-1}{8}\)
= \(-\frac{3}{4}+\frac{6}{5}-\frac{-1}{8}=\frac{-30}{40}+\frac{48}{40}+\frac{-6}{40}=\frac{12}{40}=\frac{3}{10}\)
CHÚC BN HK TỐT #