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Câu 1:
a)\(\frac{3}{4}-0,25-\left[\frac{7}{3}+\left(-\frac{9}{2}\right)\right]-\frac{5}{6}\)
\(=\frac{3}{4}-\frac{1}{4}-\frac{14}{6}+\frac{27}{6}-\frac{5}{6}\)
\(=\frac{1}{2}-\frac{4}{3}\)
\(=-\frac{5}{6}\)
b)\(7+\left(\frac{7}{12}-\frac{1}{2}+3\right)-\left(\frac{1}{12}+5\right)\)
\(=7+\frac{1}{12}+3-\frac{1}{12}-5\)
\(=5\)
Câu 2:
\(\frac{3}{4}-\frac{5}{6}\le\frac{x}{12}< 1-\left(\frac{2}{3}-\frac{1}{4}\right)\)
\(-\frac{1}{12}\le\frac{x}{12}< 1-\frac{5}{12}\)
\(-\frac{1}{12}\le\frac{x}{12}< \frac{7}{12}\)
Vậy -1\(\le\)x<7
\(A=\left(0,25\right)^{-1}.\left(\frac{1}{4}\right)^{-2}.\left(\frac{4}{3}\right)^2.\left(\frac{5}{4}\right)^{-1}.\left(\frac{2}{3}\right)^{-3}\)
\(\Rightarrow A=4^1.4^2.\frac{16}{9}.\frac{4}{5}\frac{27}{8}\)
\(\Rightarrow A=\frac{64}{1}.\frac{16}{9}.\frac{4}{5}.\frac{27}{8}\)
\(\Rightarrow A=\frac{1536}{5}\)
Vậy \(A=\frac{1536}{5}\)
\(25\cdot\left(-\frac{1}{3}\right)^3+\frac{1}{5}-2\cdot\left(-\frac{1}{2}\right)^2-\frac{1}{2}\)
\(=25\cdot\frac{-1}{27}+\frac{1}{5}-2\cdot\frac{\left(-1\right)^2}{2^2}-\frac{1}{2}\)
\(=\frac{-25}{27}+\frac{1}{5}-2\cdot\frac{1}{4}-\frac{1}{2}\)
\(=\frac{-98}{135}-\frac{1}{2}-\frac{1}{2}=\frac{-233}{135}\)
\(\left(-2\right)\left(\frac{3}{4}-0,25\right):\left(2\frac{1}{4}-1\frac{1}{6}\right)\)
\(=\left(-2\right)\left(\frac{3}{4}-\frac{25}{100}\right):\left(\frac{9}{4}-\frac{7}{6}\right)\)
\(=\left(-2\right)\left(\frac{75}{100}-\frac{25}{100}\right):\left(\frac{27}{12}-\frac{14}{12}\right)\)
\(=\left(-2\right)\cdot\frac{50}{100}:\frac{13}{12}\)
\(=\left(-2\right)\cdot\frac{1}{2}:\frac{13}{12}\)
\(=-1\cdot\frac{12}{13}=-\frac{12}{13}\)
\(A=\left(3\dfrac{1}{3}+2,5\right):\left(3\dfrac{1}{6}-4\dfrac{1}{5}\right)-\dfrac{11}{31}\\ =\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{31}{11}\\ =\left(\dfrac{30}{6}+\dfrac{15}{6}\right):\left(\dfrac{95}{30}-\dfrac{126}{30}\right)-\dfrac{31}{11}\\ =\dfrac{45}{6}:\dfrac{-21}{30}-\dfrac{31}{11}\\ =\dfrac{15}{2}\times\dfrac{-10}{7}-\dfrac{31}{11}=-\dfrac{75}{7}-\dfrac{31}{11}=-\dfrac{825}{77}-\dfrac{217}{77}=\dfrac{-1042}{77}\)
\(B=\left(-6\right).10:\left[-0,25+\dfrac{1}{2}:\left(-2\right)\right]+1\dfrac{3}{4}\\ =-60:\left(\dfrac{-1}{4}+\dfrac{1}{2}.\dfrac{-1}{2}\right)+1\dfrac{3}{4}\\ =-60:\left(\dfrac{-1}{4}+\dfrac{-1}{4}\right)+1\dfrac{3}{4}\\ =-60:\left(\dfrac{-1}{2}\right)+1\dfrac{3}{4}=120+1\dfrac{3}{4}=121\dfrac{3}{4}\)
\(\left(\frac{1}{9}\right)^{2015}.9^{2015}-96^2:24^2=1^{2015}-4^2=1-16=-15\)
\(16\frac{2}{7}:\left(\frac{-3}{5}\right)-28\frac{2}{7}:\left(\frac{-3}{5}\right)=\left(16\frac{2}{7}-28\frac{2}{7}\right):\left(\frac{-3}{5}\right)=-12.\frac{-5}{3}=20\)
\(\left(-2\right)^3.\left(\frac{3}{4}-0,25\right):\left(2\frac{1}{4}-1\frac{1}{6}\right)=-8.\frac{1}{2}:\frac{13}{12}=-8.\frac{1}{2}.\frac{12}{13}=\frac{-48}{13}\)
