\(=\frac{2^{18}.2^7.3^{14}.3^3+3^{15}.2^{15}}{2^{10}.2^{15}.3^{15}+3^{14}.3.5.2^{26}}=\frac{2^{25}.3^{17}+3^{15}.2^{15}}{2^{25}.3^{15}+3^{15}.2^{26}.5}=\frac{2^{15}.3^{15}\left(2^{10}.3^2+1\right)}{2^{25}.3^{15}\left(1+2.5\right)}\)

\(=\frac{2^{10}.3^2+1}{2^{10}\left(1+2.5\right)}=\frac{2^{10}.3^2+1}{11.2^{10}}\)

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:\(D=\frac{2^{25}.3^{15}+3^{15}.5.2^{26}}{2^{25}.3^{17}+3^{15}.2^{25}}=\frac{2^{25}3^{15}\left(1+5.2\right)}{2^{25}3^{15}\left(3^2+1\right)}=\frac{11}{10}\)

A=5 296 163 666_x10¹⁵

1+2+3+4+5+6+7+8+9+10=55

11+12+13+14+15+16+17+18+19+20=155

1+2+3+4+5+6+7+8+9+10+11+12+13+14 +15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30-50-53=362