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a.\(\frac{11^4.11^5}{11^4-11^5}:\frac{9^8.3-9^9}{9^8.5-9^8.7}:\frac{10^5-10^5.3}{10^5.11}\)
=\(\frac{11^4.11^5}{11^4-11^4.11}:\frac{9^8.3-9^8.9}{9^8.5-9^8.7}:\frac{10^5-10^5.3}{10^5.11}\)
=\(\frac{11^4.11^5}{11^4.\left(1-11\right)}:\frac{9^8.\left(3-9\right)}{9^8.\left(5-7\right)}:\frac{10^5.\left(1-3\right)}{10^5.11}
\)
=\(\frac{11^4.11^5}{11^4.\left(-10\right)}:\frac{9^8.\left(-6\right)}{9^8.\left(-2\right)}:\frac{10^5.\left(-2\right)}{10^5.11}\)
=\(\frac{11^5}{-10}:3:\frac{-2}{11}\)
=\(\frac{11^5}{-10}.\frac{1}{3}.\frac{-11}{2}\)
Chiều mình làm tiếp cho nha!👋
I don't now
sorry
...................
nha
\(A=\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^2}{3^{15}}=9\)
\(B=\frac{45^{10}.5^{10}}{75^{10}}=\frac{5^{10}.3^{20}.5^{10}}{5^{20}.3^{10}}=3^{10}\)
\(C=\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,4\right)^5.\left(0,2\right)^5}{0,4^6}=\frac{0,2^5}{0,2^2}=0,2^3\)
\(D=\frac{8^{10}+4^{10}}{8^{11}+4^{11}}=\frac{4^{10}\left(2^{10}+1\right)}{4^{11}\left(2^{11}+1\right)}=\frac{2^{10}+1}{2^{13}+1}\)
a) \(4\frac{5}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)=\frac{41}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)\)
\(=\frac{41}{9}\cdot\left(-\frac{7}{5}\right)+\frac{49}{9}\cdot\left(-\frac{7}{5}\right)=\left(\frac{41}{9}+\frac{49}{9}\right)\cdot\left(-\frac{7}{5}\right)=10\cdot\left(-\frac{7}{5}\right)=-14\)
b) \(\left(\frac{-3}{5}+\frac{4}{9}\right):\frac{7}{11}+\left(\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{4}{9}+\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{-2}{5}+\frac{4}{9}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(-1+1\right):\frac{7}{11}=0\cdot\frac{11}{7}=0\)
c) \(\left(\frac{3}{4}\right)^4\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\right)^2\cdot\left(\frac{3}{4}\right)^2\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{8}{9}\right)^2\)
\(=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
d) \(\left(-\frac{3}{5}\right)^6\cdot\left(-\frac{5}{3}\right)^5=\left(-\frac{3}{5}\right)^5\cdot\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)^5=\left[\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)\right]^5\cdot\left(-\frac{3}{5}\right)\)
\(=1^5\cdot\left(-\frac{3}{5}\right)=1\cdot\left(-\frac{3}{5}\right)=-\frac{3}{5}\)
e) \(\frac{8^{14}}{4^4\cdot64^5}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^4\cdot\left(2^6\right)^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)
f) \(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{\left(3^2\right)^{10}\cdot\left(3^3\right)^7}{\left(3^4\right)^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)
3) C thiếu đề
4) \(D=\frac{1}{9}-\left|\frac{-5}{23}\right|-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{4}-\frac{7}{30}\)
\(D=\frac{1}{9}-\frac{5}{23}+\frac{5}{23}-\frac{1}{9}-\frac{25}{7}+\frac{50}{4}-\frac{7}{30}\)
\(D=\frac{1}{9}-\frac{1}{9}-\frac{5}{23}+\frac{5}{23}+\frac{-25}{7}+\frac{50}{4}-\frac{7}{30}\)
\(D=0+0+\frac{125}{14}-\frac{7}{30}\)
\(D=\frac{913}{105}\)
Câu 1:\(\frac{45^{10}.5^{10}}{75^{10}}\) = \(\frac{\left(5.9\right)^{10}.5^{10}}{\left(5.5.3\right)^{10}}\) = \(\frac{5^{10}.9^{10}.5^{10}}{5^{10}.5^{10}.3^{10}}\) = \(\frac{9^{10}}{3^{10}}\) = \(\frac{3^{10}.3^{10}}{3^{10}}\) = \(3^{10}\) = 59049
Câu 2:\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}\) = \(\frac{\left(0,4.2\right)^5}{\left(0,4\right)^6}\) = \(\frac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}\) = \(\frac{2^5}{0,4}\) = \(\frac{32}{0,4}\) = 80
Câu 3:\(\frac{2^{15}.9^4}{6^3.8^3}\) = \(\frac{2^{15}.3^8}{2^{12}.3^3}\) = \(\frac{2^3.3^5}{1.1}\) = \(\frac{8.243}{1}\) = 1944
Câu 4: \(\frac{8^{10}+4^{10}}{8^4+4^{11}}\) = \(\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}\) = \(\frac{2^{30}+2^{20}}{2^{12}+2^{22}}\) = \(\frac{2^{20}.2^{10}+2^{20}}{2^{12}+2^{12}.2^{10}}\) = \(\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}\) = \(\frac{2^{20}}{2^{12}}\) = \(\frac{2^8}{1}\) = \(2^8\) = 256