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\(=\frac{2^{18}.2^7.3^{14}.3^3+3^{15}.2^{15}}{2^{10}.2^{15}.3^{15}+3^{14}.3.5.2^{26}}=\frac{2^{25}.3^{17}+3^{15}.2^{15}}{2^{25}.3^{15}+3^{15}.2^{26}.5}=\frac{2^{15}.3^{15}\left(2^{10}.3^2+1\right)}{2^{25}.3^{15}\left(1+2.5\right)}\)
\(=\frac{2^{10}.3^2+1}{2^{10}\left(1+2.5\right)}=\frac{2^{10}.3^2+1}{11.2^{10}}\)
Ta có:
\(A=\frac{2^{18}.18^7.3^3+3^{15}.2^{15}}{2^{10}.6^{15}+3^{14}.15.4^{13}}=\frac{2^{18}.\left(2.3^2\right)^7.3^3+3^{15}.2^{15}}{2^{10}.\left(2.3\right)^{15}+3^{14}.3.5.\left(2^2\right)^{13}}\)
\(=\frac{2^{18}.2^7.3^{14}.3^3+3^{15}.2^{15}}{2^{10}.2^{15}.3^{15}+3^{15}.5.2^{16}}=\frac{2^{25}.3^{17}+2^{15}.3^{15}}{2^{25}.3^{15}+3^{15}.2^{16}.5}=\frac{2^{15}.3^{15}.\left(3^2.2^{10}+1\right)}{2^{16}.3^{15}.\left(2^9+5\right)}\)
\(=\frac{3^2.2^{10}+1}{2^{10}+10}=\frac{9.1024+1}{1024+10}=\frac{9217}{1025}\)
\(=\dfrac{2^{18}\cdot3^3\cdot\left(3^2\cdot2\right)^7+3^{15}\cdot2^{15}}{2^{10}\cdot2^{15}\cdot3^{15}+3^{14}\cdot3\cdot5\cdot2^6}\)
\(=\dfrac{2^{25}\cdot3^{17}+3^{15}\cdot2^{15}}{2^{25}\cdot3^{15}+3^{15}\cdot5\cdot2^6}\)
\(=\dfrac{2^{15}\cdot3^{15}\left(2^{10}\cdot3^2+1\right)}{2^6\cdot3^{15}\left(2^{19}+5\cdot1\right)}=\dfrac{2^9\cdot9217}{524293}\)
#)Giải :
Câu 1 :
Đặt \(A=\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{27}\)
\(\Rightarrow A>\frac{1}{27}+\frac{1}{27}+...+\frac{1}{27}\)( 8 số hạng )
\(\Rightarrow A>\frac{8}{27}=\frac{8}{27}\)
\(\Rightarrow A>\frac{8}{27}\)
#~Will~be~Pens~#
Câu 1:(trội)
Ta có:\(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{27}>\frac{1}{27}+\frac{1}{27}+...+\frac{1}{27}=\frac{8}{27}\left(đpcm\right)\)
Câu 2:\(D=\frac{2^{25}.3^{15}+3^{15}.5.2^{26}}{2^{25}.3^{17}+3^{15}.2^{25}}=\frac{2^{25}3^{15}\left(1+5.2\right)}{2^{25}3^{15}\left(3^2+1\right)}=\frac{11}{10}\)
1/ \(\frac{9.5^{20}.27^9-3.9^{15}.25^9}{7.3^{29}.125^6-3.3^9.15^{19}}\)
\(=\frac{5^{20}.3^{29}-3^{31}.5^{18}}{7.3^{29}.5^{18}-3^{29}.5^{19}}=\frac{3^{29}.5^{18}.\left(25-9\right)}{3^{29}.5^{18}.\left(7-5\right)}=\frac{16}{2}=8\)
CÁC BÀI CÒN LẠI TƯƠNG TỰ HẾT NHÉ E
A/ \(\left(15-6\frac{13}{18}\right):11\frac{1}{27}-2\frac{1}{8}:1\frac{11}{40}\)
\(=\left(15-\frac{121}{18}\right):\frac{298}{27}-\frac{17}{8}:\frac{51}{40}\)
\(=\left(\frac{270}{18}-\frac{121}{18}\right):\frac{298}{27}-\frac{17}{8}:\frac{51}{40}\)
\(=\frac{149}{18}:\frac{298}{27}-\frac{17}{8}:\frac{51}{40}\)
\(=\frac{3}{4}-\frac{5}{3}\)
\(=\frac{9}{12}-\frac{20}{12}\)\(=-\frac{11}{12}\)
B/ \(\left(-3,2\right)\cdot-\frac{15}{64}+\left(0,8-2\frac{4}{15}\right):3\frac{2}{3}\)
\(=\left(-3,2\right)\cdot-\frac{15}{64}+\left(0,8-\frac{34}{15}\right):\frac{11}{3}\)
\(=-\frac{3,2}{1}\cdot-\frac{15}{64}+\left(0,8-\frac{34}{15}\right):\frac{11}{3}\)
\(=\frac{48}{64}+\left(0,8-\frac{34}{15}\right):\frac{11}{3}\)
\(=\frac{3}{4}+\left(\frac{12}{15}-\frac{34}{15}\right):\frac{11}{3}\)
\(=\frac{3}{4}+\left(-\frac{22}{15}\right):\frac{11}{3}\)
\(=\frac{3}{4}+\left(-\frac{2}{5}\right)\)
\(=\frac{15}{20}+\left(-\frac{8}{20}\right)\)
\(=\frac{7}{20}\)