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a) \(A=\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(A=\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\)
\(A=\dfrac{11.3^{29}-3^{29}.3}{2^2.3^{28}}\)
\(A=\dfrac{3^{29}.\left(11-3\right)}{2^2.3^{28}}\)
\(A=\dfrac{3^{29}.8}{2^2.3^{28}}\)
\(A=\dfrac{3.8}{4}=6\)
vậy \(A=6\)
b) \(B=\dfrac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(B=\dfrac{3^2.\left(2^2\right)^2.\left(2^{16}\right)^2}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)
\(B=\dfrac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(B=\dfrac{3^2.2^{36}}{11.2^{35}-2^{35}.2}\)
\(B=\dfrac{3^2.2^{36}}{2^{35}.\left(11-2\right)}\)
\(B=\dfrac{3^2.2^{36}}{2^{35}.9}\)
\(B=\dfrac{3^2.2}{9}\)
\(B=\dfrac{9.2}{9}\)
\(B=2\)
vậy \(B=2\)
A=11.322.37-915/(2.314)2
A=11.329-915/22.328
A=11.329-(32)15/4.328
A=11.329-330/4.328
A=11.329-329.3/4.328
A=329.(11-3)/328.4
A=329.8/328.4
A=3.8/4
A=24/4
A=6
\(=\dfrac{11\cdot3^{21}-3^{30}}{2^2\cdot3^{28}}=\dfrac{3^{21}\left(11-3^7\right)}{2^2\cdot3^{28}}=\dfrac{-2176}{2^2\cdot3^7}=\dfrac{-2176}{8748}=\dfrac{-544}{2187}\)
\(=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}=\frac{3^{29}.2^3}{2^2.3^{28}}=3.2=6.\)
\(\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{22}.3^7-\left(3^2\right)^{15}}{2^2.\left(3^{14}\right)^2}=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}\)
\(=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}=\frac{3.8}{4}=\frac{24}{4}=6\)
Đáp số: \(6\)
\(A=\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)
\(B=\frac{\left(3\cdot4\cdot2^{16}\right)^2}{11\cdot2^{13}\cdot4^{11}-16^9}=2\)
\(C=\frac{4^5\cdot9^{4-2\cdot6^9}}{2^{10}\cdot3^8+6^8\cdot20}=0\)
A=\(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=6\)
\(\dfrac{11\cdot3^7\cdot9^7-9^{15}}{\left(2\cdot3^{14}\right)^2}\)
\(=\dfrac{11\cdot3^7\cdot3^{14}-3^{30}}{2^2\cdot3^{28}}\)
\(=\dfrac{3^{21}\left(11-3^9\right)}{2^2\cdot3^{28}}\)
\(=\dfrac{11-3^9}{2^2\cdot3^7}\)
a, \(\left(\dfrac{-2}{3}+\dfrac{3}{7}\right)-\dfrac{5}{21}:\dfrac{4}{5}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\\ = -\dfrac{5}{21}:\dfrac{4}{5}+ \left(-\dfrac{5}{21}\right):\dfrac{4}{5}\\ =\left[-\dfrac{5}{21}+\left(-\dfrac{5}{21}\right)\right]:\dfrac{4}{5}\\ -\dfrac{10}{21}:\dfrac{4}{5}\\ =-\dfrac{25}{42}\)
b,
\(\dfrac{5}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)\\ =\dfrac{5}{9}:\dfrac{-3}{22}+\dfrac{5}{9}:-\dfrac{3}{5}\\ =\dfrac{5}{9}:\left(\dfrac{-3}{22}+-\dfrac{3}{5}\right)\\ =\dfrac{5}{9}:-\dfrac{81}{110}\\ =-\dfrac{550}{729}\)
\(A=\dfrac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^2\right)^6.3+\left(2^3\right)^4.3^5}-\dfrac{5^{10}+7^3-\left(5^2\right)^5.\left(7^2\right)^2}{(5^3).7^3+5^9.\left(7^2\right)^3}\)
\(A=\dfrac{2^4.1-2.3^3}{1.1+1.1}-\dfrac{5+1-5.1}{1.1+1.7}\)
\(A=\dfrac{2^4-54}{2}-\dfrac{1.1}{2\cdot7}\)
\(A=(2^3-54)-\dfrac{1}{14}\)
\(A=\left(-46\right)-\dfrac{1}{14}\)
A=\(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\dfrac{11.3^{29}-3^{29}.3}{2^2.3^{28}}=\dfrac{3^{29}.\left(11-3\right)}{2^2.3^{28}}=\dfrac{3^{29}.2^3}{2^2.3^{28}}=3.2=6\)