The value of \(\frac{9^4\cdot27^3}{81^4}-\frac{2^{11}\cdot4^5}{6^{10}}\) is .........
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\(A=\frac{15.3^{11}+4.27^1}{9^7}\)
\(\Rightarrow A=\frac{3.5.3^{11}+4.3^{3^1}}{\left(3^2\right)^7}\)
\(\Rightarrow A=\frac{3^{12}.5+4.3^3}{3^{14}}\)
\(\Rightarrow A=\frac{3^3.\left(5.3^8+4.3^3\right)}{3^{14}}\)
\(\Rightarrow A=\frac{32805+4}{177147}\)
\(\Rightarrow A=\frac{32809}{177147}\)
bài này không khó. Nhưng đánh máy để giải cho bạn thì thực sự khó
\(I=\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.2^{30}.3^{27}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{5.2^{30}.3^{27}-3^{30}.2^{29}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{27}.\left(5.2-3^3\right)}{2^{28}.3^{18}.\left(5.3-2.7\right)}\)
\(=\frac{2^{29}.3^{27}.-17}{2^{18}.3^{18}}\)
\(=\frac{2^9.3^9.-17}{1}\)
Ta có \(H=\frac{\left(3.4.2^{16}\right)}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3.4.2^{16}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3.2^{18}}{11.2^{35}-2^{36}}\)
\(=\frac{3.2^{18}}{2^{35}.\left(11-2\right)}\)
\(=\frac{3.2^{18}}{2^{35}.3^2}\)
\(=\frac{1}{2^{17}.3}\)
\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(=\frac{2^{19}.\left(3^3\right)^3+15.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(=\frac{2^{19}.3^9+15.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(=\frac{2^{19}.3^9+15.2^{18}.3^8}{2^{19}.3^9+2^{20}.3^{10}}\)
\(=\frac{2^{18}.3^8\left(2.3+15\right)}{2^{19}.3^9\left(1+2.3\right)}\)
\(=\frac{6+15}{2.3\left(1+6\right)}\)
\(=\frac{21}{6.7}\)
\(=\frac{21}{42}\)
\(=\frac{1}{2}\)
Có P =\(\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\dfrac{2^{19}.\left(3^3\right)^3+5.3.\left(3^2\right)^4}{\left(2.3\right)^9+\left(3.2^2\right)^{10}}\)=\(\dfrac{2^{19}.3^9+5.3.2^{18}.3^8}{3^9.2^9.2^{10}+3^{10}.\left(2^2\right)^{10}}=\dfrac{2^{19}.3^9+5.2^{18}.3^9}{3^9.2^{19}+3^{10}.2^{20}}=\dfrac{2^{18}.3^9.\left(2+5\right)}{3^9.2^{19}.\left(1+3.2\right)}=\dfrac{2^{18}.3^9.7}{3^9.2^{19}.7}\)
=\(\dfrac{1}{2}\)
\(=\frac{2^{19}3^9+3\cdot5\cdot2^{18}\cdot3^8}{2^9\cdot3^9\cdot2^{10}+4^{10}\cdot3^{10}}=\frac{2^{19}\cdot3^9+5\cdot2^{18}\cdot3^9}{2^{19}\cdot3^9+2^{20}\cdot3^{10}}=\frac{2^{18}\cdot3^9\cdot\left(2+5\right)}{2^{19}\cdot3^9\left(1+6\right)}=\frac{1}{2}\)