Bài 7:
\(a,3^7:3^5-\left(\dfrac{5}{17}\right)^0\)
\(=3^{7-5}-1\)
\(=3^2-1\)
\(=9-1=8\)
\(b,\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{1}{2}+2\right)^3\)
\(=\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{5}{2}\right)^3\)
\(=\left(\dfrac{5}{2}\right)^{13-3}\)
\(=\dfrac{5}{2}^{10}\)
\(c,8.\left(\dfrac{1}{4}\right)^3+\left(\dfrac{2}{27}\right)^0-\dfrac{1}{8}\)
\(=\dfrac{1}{8}+1-\dfrac{1}{8}\)
\(=\left(\dfrac{1}{8}-\dfrac{1}{8}\right)+1\)
\(=0+1=1\)
\(d,3.\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)
\(=\dfrac{8}{9}-1+\dfrac{4}{9}\)
\(=\dfrac{8}{9}-\dfrac{9}{9}+\dfrac{4}{9}\)
\(=\dfrac{3}{9}=\dfrac{1}{3}\)
`8`
\(a,\dfrac{8^5}{4^7}\)
\(=\dfrac{\left(2^3\right)^5}{\left(2^2\right)^7}=\dfrac{2^{15}}{2^{14}}\)
\(=2^{15}:2^{14}\)
\(=2\)
\(b,\dfrac{49^2.7^8}{98.7^9}\)
\(=\dfrac{\left(7^2\right)^2.7^8}{7^2.2.7^9}\)
\(=\dfrac{7^4.7^8}{7^2.2.7^9}\)
\(=\dfrac{7.1}{1.2.1}\)
\(=\dfrac{7}{2}\)
\(c,\dfrac{21^{30}}{63^{15}}\)
\(=\dfrac{\left(7.3\right)^{30}}{\left(7.3^2\right)^{15}}\)
\(=\dfrac{7^{30}.3^{30}}{7^{15}.3^{30}}\)
\(=7^{15}\)
\(d,\dfrac{28^9.30^9}{105^9}\)
\(=\dfrac{\left(7.4\right)^9.\left(3.5.2\right)^9}{\left(5.7.3\right)^9}\)
\(=\dfrac{7^9.2^{18}.3^9.5^9.2^9}{5^9.7^9.3^9}\)
\(=\dfrac{1.2^{18}.1.1.2^9}{1.1.1}\)
\(=2^{18}.2^9\)
\(=2^{27}\)
`7`
\(a,3^7:3^3-(\dfrac{5}{17})^0\)
\(=3^7:3^3-1\)
\(=3^2-1\)
\(=9-1\)
\(=8\)
\(b,(\dfrac{5}{2})^{13}:\left(\dfrac{1}{2}+2\right)^3\)
\(=\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{5}{2}\right)^3\)
\(=\left(\dfrac{5}{2}\right)^{13-3}\)
\(=\left(\dfrac{5}{2}\right)^{10}\)
\(c,8.\left(\dfrac{1}{4}\right)^3+\left(\dfrac{2}{27}\right)^0-\dfrac{1}{8}\)
\(=8.\dfrac{1}{64}+1-\dfrac{1}{8}\)
\(=\dfrac{1}{8}+1-\dfrac{1}{8}\)
\(=\left(\dfrac{1}{8}-\dfrac{1}{8}\right)+1\)
\(=0+1\)
\(=1\)
\(d,3.\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)
\(=3.\left(\dfrac{2}{3}\right)^3+(\dfrac{52}{3})^0+\dfrac{4}{9}\)
\(=3.\dfrac{8}{27}+1+\dfrac{4}{9}\)
\(=1.\dfrac{8}{9}+1+\dfrac{4}{9}\)
\(=\dfrac{8}{9}+1+\dfrac{4}{9}\)
\(=\left(\dfrac{8}{9}+\dfrac{4}{9}\right)+1\)
\(=\dfrac{12}{9}+\dfrac{9}{9}\)
\(=\dfrac{7}{3}\)
8.
\(a,\dfrac{8^5}{4^7}=\dfrac{\left(2^3\right)^5}{\left(2^2\right)^7}=\dfrac{2^{15}}{2^{14}}=2\)
\(b,\dfrac{49^2.7^8}{98.7^9}=\dfrac{\left(7^2\right)^2.7^8}{2.7^2.7^9}=\dfrac{7^4.7^8}{2.7^{11}}=\dfrac{7^{12}}{2.7^{11}}=\dfrac{7}{2}\)
\(c,\dfrac{21^{30}}{63^{15}}=\dfrac{21^{30}}{21^{15}.3^{15}}=\dfrac{21^{15}}{3^{15}}=\dfrac{\left(3.7\right)^{15}}{3^{15}}=\dfrac{3^{15}.7^{15}}{3^{15}}=7^{15}\)
\(d,\dfrac{28^9.30^9}{105^9}=\dfrac{\left(2^2.7\right)^9.\left(5.2.3\right)^9}{\left(3.7.5\right)^9}=\dfrac{2^{18}.7^9.5^9.2^9.3^9}{3^9.7^9.5^9}=2^{18}.2^9=2^{18+9}=2^{27}\)
9.
a) \(\dfrac{\left(-3\right)^{10}.15^5}{25^3.\left(-9\right)^7}=\dfrac{\left(-1.3\right)^{10}.\left(3.5\right)^5}{\left(5^2\right)^3.\left(-1.3^2\right)^7}=\dfrac{\left(-1\right)^{10}.3^{10}.3^5.5^5}{5^6.\left(-1\right)^7.3^{14}}=\dfrac{3^{15}.5^5}{5^6.3^{14}.\left(-1\right)}=\dfrac{3}{-5}=\dfrac{-3}{5}\)
\(b,2^3+3\left(\dfrac{1}{9}\right)^0-2^{-2}.4+\left[\left(-2\right)^2:\dfrac{1}{2}\right].8\)
\(=8+3.1-\dfrac{1}{4}.4+\left[4:\dfrac{1}{2}\right].8\)
\(=8+3-\dfrac{1}{4}.4+8.8\)
\(=8+3-1+64\)
\(=74\)
`9`
\(a,\dfrac{\left(-3\right)^{10}.15^5}{25^3.\left(-9\right)^7}\)
\(=\dfrac{\left(-3\right)^{10}.\left(3.5\right)^5}{\left(5^2\right)^3.\left(-3^2\right)^7}\)
\(=\dfrac{\left(-3\right)^{10}.3^5.5^5}{5^6.\left(-3\right)^{14}}\)
\(=\dfrac{1.3^5.1}{5.3^4}\)
\(=\dfrac{3}{5}\)
\(b,2^3+3\left(\dfrac{1}{9}\right)^0-2^{-2}.4+\left[\left(-2\right)^2:\dfrac{1}{2}\right].8\)
\(=8+3.1-\dfrac{1}{4}.4+\left[4:\dfrac{1}{2}\right].8\)
\(=8+3.1-\dfrac{1}{4}.4+8.8\)
\(=8+3-1+64\)
\(=11-1+64\)
\(=10+64\)
\(=74\)