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\(B=\frac{3^9-2^3\cdot3^7+2^{10}\cdot3^2-2^{13}}{3^{10}-2^2\cdot3^7+2^{10}\cdot3^3-2^{12}}\)
\(B=\frac{1-2\cdot1+1\cdot1-2}{3-1\cdot1+1\cdot3-1}\)
\(B=\frac{1-2+1-2}{3-1+3-1}\)
\(B=\frac{-1+\left(-1\right)}{2+2}\)
\(B=\frac{-2}{4}\)
\(\Rightarrow B=\frac{-1}{2}\)
Ta có :
\(A=\frac{3^9-2^3.3^7+2^{10}.3^2-2^{13}}{3^{10}-2^2.3^7+2^{10}.3^3-2^{12}}\)
\(A=\frac{3^7\left(3^2-2^3\right)+2^{10}\left(3^2-2^3\right)}{3^7\left(3^3-2^2\right)+2^{10}\left(3^3-2^2\right)}\)
\(A=\frac{\left(3^2-2^3\right)\left(3^7+2^{10}\right)}{\left(3^3-2^2\right)\left(3^7+2^{10}\right)}\)
\(A=\frac{3^2-2^3}{3^3-2^2}\)
\(A=\frac{9-8}{27-4}\)
\(A=\frac{1}{23}\)
Vậy \(A=\frac{1}{23}\)
Chúc bạn học tốt ~
mk doan la` de sai, sua: \(\frac{3^9-2^3.3^7+2^{10}.3^2-2^{13}}{3^{10}-2^2.3^7+2^{10}.3^3-2^{12}}\)
\(=\frac{3^7.\left(3^2-2^3\right)+2^{10}.\left(3^2-2^3\right)}{3^7.\left(3^3-2^2\right)+2^{10}.\left(3^3-2^2\right)}=\frac{3^7+2^{10}}{\left(3^7+2^{10}\right).24}=\frac{1}{24}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^4.2^3.7^3}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^4.7^3\left(5^5+2^3\right)}\)
\(=\frac{1}{6}+\frac{93750}{3133}\)
