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\(\frac{11.3^{25}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{32}-\left(3^2\right)^{15}}{2^2.\left(3^{14}\right)^2}=\frac{11.3^{32}-3^{30}}{4.3^{28}}=\frac{3^{30}.\left(11.3^2-1\right)}{4.3^{28}}=\frac{3^2.98}{4}=\frac{9.98}{4}=\frac{882}{4}=\frac{441}{2}\)
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\(A=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right).2}\)
\(A=\frac{11.3^{22+7}-\left(3^2\right)^{15}}{2^2.\left(3^{14}\right)^2}\)
\(A=\frac{11.3^{29}-3^{30}}{4.3^{28}}\)
\(A=\frac{11.3^{29}-3^{29}.3}{4.3^{28}}\)
\(A=\frac{3^{29}.\left(11-3\right)}{3^{28}.4}\)
\(A=\frac{3^{28}.3.8}{3^{28}.4}\)
\(A=\frac{3^{28}.3.4.2}{3^{28}.4}\)
\(A=6\)
\(A=3.2\)
Vậy : \(A=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=6\)
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\(\frac{11.3^{29}-\left(3.3\right)^{15}}{2^2.3^{28}}\)=\(\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{3^{29}.\left(11-3\right)}{3^{28}.2^2}=\frac{3.8}{2^2}=3.2\)=6
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\(=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{11.3^{29}-3^{29}.3}{2^2.3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}=\frac{3^{29}.2^3}{2^2.3^{28}}=\frac{3.2}{1.1}=6\)
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B = ( 2 - 4 - 6 + 8 ) + ( 10 - 12 - 14 + 16 ) +......+ ( 98 - 100 - 102 + 104 )
B = 0 + 0 +....+ 0 ( 52 số 0 )
B = 0
A=1+(3-5)+(7-9)+...+(2015-2017)
A=1+(-2)+(-2)+...+(-2)
A=(-2)*1008+1
A=(-2014)+1
A=-2015
\(B=\frac{2^{10}\cdot55-2^{10}}{2^8\cdot27}=\frac{2^{10}\left(55-1\right)}{2^8\cdot3^3}=\frac{2^{10}\cdot2\cdot3^3}{2^8\cdot3^3}=\frac{2^{11}\cdot3^3}{2^8\cdot3^3}=2^3=8.\)
\(C=\frac{\left(3\cdot4\cdot24\right)^2}{5\cdot2^5\cdot4^2-16^2}=\frac{288^2}{5\cdot16\cdot2^3.16-16^2}=\frac{2^5\cdot3^2}{16^2\left(5\cdot2^3-1\right)}=\frac{16\cdot2\cdot9}{16\cdot16\left(5\cdot8-1\right)}\)
\(=\frac{16\cdot2\cdot9}{16\cdot16\cdot39}=\frac{16\cdot2\cdot9}{16\cdot2^4\cdot13\cdot3}=\frac{1\cdot3}{2^3\cdot13\cdot1}=\frac{3}{8\cdot13\cdot1}=\frac{3}{104}\)
\(D=\frac{11\cdot3^{22}\cdot3^7-9^{14}}{\left(2\cdot3^{14}\right)^2}=\frac{11\cdot3^{30}-3^{28}}{2^2\cdot3^{28}}=\frac{3^{28}\left(11\cdot3^2-1\right)}{4\cdot3^{38}}=\frac{3^{28}\left(11\cdot9-1\right)}{4\cdot3^{28}}=\frac{99-1}{4}=\frac{98}{4}=24,5\)
cho cách là phân số đấy