tinh
A=\(\frac{11×3^{22}×3^7-9^{15}}{\left(2×3^{14}\right)^2}\)
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\(A=\frac{3^7\cdot17-3^9}{2^3\cdot3^5}=\frac{3^7\left(17-3^2\right)}{2^3\cdot3^5}=\frac{3^7\cdot2^3}{2^3\cdot3^5}=9\)
\(B=\frac{3^2\cdot4^2\cdot2^{32}}{11\cdot2^{13}\cdot4^{11}-16^9}=\frac{3^2\cdot2^{36}}{2^{35}\cdot11-2^{36}}=\frac{3^2\cdot2^{36}}{2^{35}\left(11-2\right)}=\frac{3^2\cdot2^{36}}{2^{35}\cdot3^2}=2\)
\(\frac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2\cdot3^{28}}=\frac{3^{29}\cdot8}{2^2\cdot3^{28}}=6\)
\(\left(\frac{-2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(\frac{-1}{3}+\frac{4}{7}\right):\frac{4}{5}\)
\(=\left(\frac{-2}{3}+\frac{3}{7}+\frac{-1}{3}+\frac{4}{7}\right):\frac{4}{5}\)
\(=\left[\left(\frac{-2}{3}+\frac{-1}{3}\right)+\left(\frac{3}{7}+\frac{4}{7}\right)\right]:\frac{4}{5}\)
\(=\left[\left(-1+1\right)\right]:\frac{4}{5}\)
\(=0:\frac{4}{5}=0\)
~ Hok tốt ~
a, A = \(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)
\(A=\frac{2^{10}\left(13+65\right)}{2^8.2^2.26}=\frac{2^{10}.78}{2^{10}.26}=\frac{78}{26}=3\)
Vậy A = 3
b, \(B=\frac{72^3.54^2}{108^4}=\frac{72^3.54^2}{\left(54.2\right)^4}=\frac{72^3.54^2}{54^4.2^4}=\frac{72^3}{54^2.2^4}=\frac{\left(8.9\right)^3}{\left(6.9\right)^2.2^4}\)
\(=\frac{\left(2^3\right)^3.9^3}{6^2.9^2.2^4}=\frac{2^9.9^3}{2^2.3^2.9^2.2^4}=\frac{2^9.9^3}{2^6.9^3}=\frac{2^9}{2^6}=2^3=8\)
Vậy B = 8
c, \(C=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}.3^{30}}{2^2.3^{28}}=\frac{11.3^{29}.3.3^{29}}{2^2.3^{28}}=\frac{\left(11-3\right)3^{29}}{2^2.3^{28}}\)
\(=\frac{2^3.3^{29}}{2^2.3^{28}}=2.3=6\)
Vậy C = 6
d, \(D=\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}=\frac{\left(3.2^{18}\right)^2}{11.2^{35}-\left(2^4\right)^9}=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}=\frac{3^2.2^{36}}{\left(11-2\right)2^{35}}=\frac{3^2.2}{9}=2\)
Vậy D = 2