Tính:
a) \(\dfrac{2}{3}\times\dfrac{5}{8}+\dfrac{7}{4}\) b) \(\dfrac{8}{5}:\left(\dfrac{4}{3}-\dfrac{5}{6}\right)\) c) \(\dfrac{3}{4}\times\dfrac{1}{5}-\dfrac{1}{10}\)
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a) $\frac{2}{5} \times \frac{3}{8} \times \frac{3}{4} = \frac{{2 \times 3 \times 3}}{{5 \times 8 \times 4}} = \frac{{18}}{{160}} = \frac{9}{{80}}$
b) $\frac{1}{3} \times \frac{1}{6} \times \frac{1}{9} = \frac{{1 \times 1 \times 1}}{{3 \times 6 \times 9}} = \frac{1}{{162}}$
c) $\frac{3}{4}:\frac{1}{5}:\frac{7}{8} = \frac{3}{4} \times \frac{5}{1} \times \frac{8}{7} = \frac{{3 \times 5 \times 8}}{{4 \times 1 \times 7}} = \frac{{120}}{{28}} = \frac{{30}}{7}$
d) $\frac{3}{5}:\frac{1}{5}:\frac{3}{8} = \frac{3}{5} \times \frac{5}{1} \times \frac{8}{3} = \frac{{3 \times 5 \times 8}}{{5 \times 1 \times 3}} = 8$
a) $\frac{5}{2} \times \frac{4}{3} + \frac{1}{3} = \frac{{10}}{3} + \frac{1}{3} = \frac{{11}}{3}$
b) $\frac{7}{3} - \frac{2}{3}:\frac{5}{7} = \frac{7}{3} - \frac{2}{3} \times \frac{7}{5} = \frac{7}{3} - \frac{{14}}{{15}} = \frac{{35}}{{15}} - \frac{{14}}{{15}} = \frac{{21}}{{15}} = \frac{7}{5}$
c) $\frac{3}{4} \times \left( {\frac{5}{2} - \frac{3}{2}} \right) = \frac{3}{4} \times 1 = \frac{3}{4}$
a: \(A=\dfrac{9^4}{3^2}=\dfrac{\left(3^2\right)^4}{3^2}=\dfrac{3^8}{3^2}=3^6\)=729
b: \(B=81\left(\dfrac{5}{3}\right)^4=81\cdot\dfrac{5^4}{3^4}=\dfrac{81}{3^4}\cdot5^4=5^4=625\)
c: \(C=\left(\dfrac{4}{7}\right)^{-4}\cdot\left(\dfrac{2}{7}\right)^3\)
\(=\left(\dfrac{7}{4}\right)^4\cdot\left(\dfrac{2}{7}\right)^3\)
\(=\dfrac{7^4}{4^4}\cdot\dfrac{2^3}{7^3}\)
\(=\dfrac{2^3}{4^4}\cdot7\)
\(=\dfrac{2^3}{2^8}\cdot7=\dfrac{7}{2^5}=\dfrac{7}{32}\)
d: \(D=7^{-6}\cdot\left(\dfrac{2}{3}\right)^0\left(\dfrac{7}{5}\right)^6\)
\(=7^{-6}\left(\dfrac{7}{5}\right)^6\)
\(=\dfrac{1}{7^6}\cdot\dfrac{7^6}{5^6}=\dfrac{1}{5^6}=\dfrac{1}{15625}\)
e: \(E=8^3:\left(\dfrac{2}{3}\right)^5\cdot\left(\dfrac{1}{3}\right)^2\)
\(=2^6:\dfrac{2^5}{3^5}\cdot\dfrac{1}{3^2}\)
\(=2^6\cdot\dfrac{3^5}{2^5}\cdot\dfrac{1}{3^2}\)
\(=\dfrac{2^6}{2^5}\cdot\dfrac{3^5}{3^2}=3^3\cdot2=54\)
f: \(F=\left(\dfrac{7}{9}\right)^{-2}\cdot\left(\dfrac{1}{\sqrt{3}}\right)^8\)
\(=\left(\dfrac{9}{7}\right)^2\cdot\left(\dfrac{1}{3}\right)^4\)
\(=\dfrac{9^2}{7^2}\cdot\dfrac{1}{3^4}=\dfrac{9^2}{3^4}\cdot\dfrac{1}{7^2}=\dfrac{81}{81}\cdot\dfrac{1}{49}=\dfrac{1}{49}\)
g: \(G=\left(-\dfrac{4}{5}\right)^{-2}\cdot\left(\dfrac{2}{5}\right)^2\cdot\left(\sqrt{2}\right)^3\)
\(=\left(-\dfrac{5}{4}\right)^2\cdot\left(\dfrac{2}{5}\right)^2\cdot2\sqrt{2}\)
\(=\dfrac{25}{16}\cdot\dfrac{4}{25}\cdot2\sqrt{2}=\dfrac{4}{16}\cdot2\sqrt{2}=\dfrac{8\sqrt{2}}{16}=\dfrac{\sqrt{2}}{2}\)
a: \(\dfrac{-1}{2}+\dfrac{5}{6}+\dfrac{1}{3}\)
\(=\dfrac{-3}{6}+\dfrac{5}{6}+\dfrac{2}{6}\)
\(=\dfrac{4}{6}=\dfrac{2}{3}\)
b: \(\dfrac{-3}{8}+\dfrac{7}{4}-\dfrac{1}{12}\)
\(=\dfrac{-9}{24}+\dfrac{42}{24}-\dfrac{2}{24}\)
\(=\dfrac{31}{24}\)
c: \(\dfrac{3}{5}:\left(\dfrac{1}{4}\cdot\dfrac{7}{5}\right)=\dfrac{3}{4}:\dfrac{7}{20}=\dfrac{3}{4}\cdot\dfrac{20}{7}=\dfrac{15}{7}\)
d: \(\dfrac{10}{11}+\dfrac{4}{11}:4-\dfrac{1}{8}\)
\(=\dfrac{10}{11}+\dfrac{1}{11}-\dfrac{1}{8}=\dfrac{7}{8}\)
a: \(\dfrac{3}{9}\times\dfrac{5}{4}=\dfrac{5}{4}\times\dfrac{1}{3}=\dfrac{5\times1}{4\times3}=\dfrac{5}{12}\)
b: \(\dfrac{10}{15}\times\dfrac{3}{5}=\dfrac{3}{5}\times\dfrac{2}{3}=\dfrac{3\times2}{5\times3}=\dfrac{2}{5}\)
c: \(\dfrac{5}{8}\times\dfrac{4}{12}=\dfrac{5}{8}\times\dfrac{1}{3}=\dfrac{5}{3\times8}=\dfrac{5}{24}\)
d: \(\dfrac{9}{27}\times\dfrac{3}{21}=\dfrac{1}{7}\times\dfrac{1}{3}=\dfrac{1\times1}{7\times3}=\dfrac{1}{21}\)
a) \(\dfrac{2}{3}\times\dfrac{4}{5}=\dfrac{4}{5}\times\dfrac{2}{3}\)
b) \(\left(\dfrac{1}{3}\times\dfrac{2}{5}\right)\times\dfrac{3}{4}=\dfrac{1}{3}\times\left(\dfrac{2}{5}\times\dfrac{3}{4}\right)\)
c) \(\left(\dfrac{1}{3}-\dfrac{2}{15}\right)\times\dfrac{3}{4}=\dfrac{1}{3}\times\dfrac{3}{4}+\dfrac{2}{15}\times\dfrac{3}{4}\)
a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)
\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)
\(=\left(-\dfrac{1}{2}\right)2+1\)
\(=-1+1\)
\(=0\)
@Trịnh Thị Thảo Nhi
a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1
=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1
=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1
=(−12)2+1=(−12)2+1
=−1+1=−1+1
=0=0
a) $\frac{3}{2} \times \frac{5}{8} + \frac{7}{4} = \frac{{15}}{{16}} + \frac{7}{4} = \frac{{15}}{{16}} + \frac{{28}}{{16}} = \frac{{43}}{{16}}$
b) $\frac{8}{5}:\left( {\frac{4}{3} - \frac{5}{6}} \right) = \frac{8}{5}:\left( {\frac{8}{6} - \frac{5}{6}} \right) = \frac{8}{5}:\frac{1}{2} = \frac{8}{5} \times 2 = \frac{{16}}{5}$
c) $\frac{3}{4} \times \frac{1}{5} - \frac{1}{{10}} = \frac{3}{{20}} - \frac{1}{{10}} = \frac{3}{{20}} - \frac{2}{{20}} = \frac{1}{{20}}$