tính một cách hợp lí :
\((\frac {-1}{3})^7×3^7+(0,125)^3×8^3+(\frac{1}{5})^3×10^3 \)
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a)
\(\begin{array}{l}\frac{{ - 3}}{{10}} - 0,125 + \frac{{ - 7}}{{10}} + 1,125 \\= \left( {\frac{{ - 3}}{{10}} + \frac{{ - 7}}{{10}}} \right) + \left( {1,125 - 0,125} \right)\\ = - 1 + 1 \\= 0\end{array}\)
b)
\(\begin{array}{l}\frac{{ - 8}}{3}.\frac{2}{{11}} - \frac{8}{3}:\frac{{11}}{9} \\= \frac{8}{3}.\frac{{ - 2}}{{11}} - \frac{8}{3}.\frac{9}{{11}}\\ = \frac{8}{3}.\left( {\frac{{ - 2}}{{11}} - \frac{9}{{11}}} \right)\\ =\frac{{ - 8}}{3}.\frac{-11}{11}\\= \frac{8}{3}.\left( { - 1} \right) \\= \frac{{ - 8}}{3}\end{array}\)
Ta có: \(A=\frac{\frac{3}{11}+1-\frac{3}{7}}{3+\frac{9}{11}-\frac{9}{7}}-\frac{\frac{1}{3}+0,25-\frac{1}{5}+0,125}{\frac{7}{6}+\frac{7}{8}-0,7+\frac{7}{16}}\)
\(=\frac{3\left(\frac{1}{11}+\frac{1}{3}-\frac{1}{7}\right)}{9\left(\frac{1}{3}+\frac{1}{11}-\frac{1}{7}\right)}-\frac{2\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}+\frac{1}{16}\right)}{7\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}+\frac{1}{16}\right)}\)
\(=\frac{3}{9}-\frac{2}{7}=\frac{1}{3}-\frac{2}{7}=\frac{7}{21}-\frac{6}{21}=\frac{1}{21}\)
Vậy \(A=\frac{1}{21}\)
\(\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}+\frac{1}{6}.\frac{1}{7}+\frac{1}{7}.\frac{1}{8}+\frac{1}{8}.\frac{1}{9}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}\)
\(=\frac{9}{18}-\frac{2}{18}\)
\(=\frac{7}{18}\)
\(\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}+\frac{1}{6}.\frac{1}{7}+\frac{1}{7}.\frac{1}{8}+\frac{1}{8}.\frac{1}{9}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}\)
\(=\frac{7}{18}\)
Chúc bạn học tốt !!!!
\(\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{8}-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}{\frac{3}{4}+\frac{1}{2}-\frac{3}{10}}\)\(=\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{3\left(\frac{1}{8}-\frac{1}{5}+\frac{1}{7}\right)}+\frac{60.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}\right)}{60.\left(\frac{3}{4}+\frac{1}{2}-\frac{3}{10}\right)}\)
\(=\frac{1}{3}+\frac{30+20-12}{45+30-18}\)\(=\frac{1}{3}+\frac{38}{57}=1\)
\(\frac{0,125-\frac{1}{5}+\frac{1}{7}}{0,375-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{1}{2}+\frac{1}{3}-0,2}{\frac{3}{4}+0,5-\frac{3}{10}}\)
= \(\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{8}-\frac{3}{5}+\frac{3}{7}}+\frac{\frac{2}{4}+\frac{2}{6}-\frac{2}{10}}{\frac{3}{4}+\frac{3}{6}-\frac{3}{10}}\)
= \(\frac{\frac{1}{8}-\frac{1}{5}+\frac{1}{7}}{3.\left(\frac{1}{8}-\frac{1}{5}+\frac{1}{7}\right)}+\frac{2.\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{10}\right)}{3.\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{10}\right)}\)
= \(\frac{1}{3}+\frac{2}{3}\)
= \(\frac{3}{3}\)
= 1
\(=\frac{\left(-1\right)^7}{3^7}.3^7+\left(\frac{1}{8}\right)^3.8^3+\left(\frac{1}{5}\right)^3.\left(2.5\right)^3\)
\(=\left(-1\right)^7+\frac{1^3}{8^3}.8^3+\frac{1^3}{5^3}.2^3.5^3\)
\(=-1+1+2^3=2^3=8\)