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

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

Ta có:

\(A=\frac{2^{18}.18^7.3^3+3^{15}.2^{15}}{2^{10}.6^{15}+3^{14}.15.4^{13}}=\frac{2^{18}.\left(2.3^2\right)^7.3^3+3^{15}.2^{15}}{2^{10}.\left(2.3\right)^{15}+3^{14}.3.5.\left(2^2\right)^{13}}\)

\(=\frac{2^{18}.2^7.3^{14}.3^3+3^{15}.2^{15}}{2^{10}.2^{15}.3^{15}+3^{15}.5.2^{16}}=\frac{2^{25}.3^{17}+2^{15}.3^{15}}{2^{25}.3^{15}+3^{15}.2^{16}.5}=\frac{2^{15}.3^{15}.\left(3^2.2^{10}+1\right)}{2^{16}.3^{15}.\left(2^9+5\right)}\)

\(=\frac{3^2.2^{10}+1}{2^{10}+10}=\frac{9.1024+1}{1024+10}=\frac{9217}{1025}\)

lớp 1 mà cậu

4.24.5²-(3³.18+3³.12)

=4.24.25-[27.(18+12)]

=(4.25).24-[27.30]

=100.24-810

=2400-810

=1590

4.24.5^2-(3^3.18+3^3.12)

31.15.7^2.4-31.49.40

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

1+3+5+7+9+11+13+15+19

2+6+10+14+22+23+26+34

5+8+11+14+17+20+23+26+29

1+6+11+16+21+26+31+36+41+47+51

10+13+16+19+22+25+28+31+34+37+40

5+7+9+11+13+15+17+3+8+13+18+23+28

4+7+10+13+16+19+5+9+13+17+21+25 = 1590

#HT#

\(=\dfrac{2^{18}\cdot3^{14}\cdot3^3\cdot2^7+3^{15}\cdot2^{15}}{2^{10}\cdot2^{15}\cdot3^{15}+3^{14}\cdot3\cdot5\cdot2^{26}}\)

\(=\dfrac{2^{25}\cdot3^{17}+3^{15}\cdot2^{15}}{2^{25}\cdot3^{15}+3^{15}\cdot5\cdot2^{26}}\)

\(=\dfrac{2^{15}\cdot3^{15}\left(2^{10}\cdot3^2+1\right)}{2^{25}\cdot3^{15}\left(1+5\cdot2\right)}=\dfrac{1}{1024}\cdot\dfrac{9217}{11}=\dfrac{9217}{11264}\)

tk

thực hiện phép tính;        a.\(\frac{2^{18}.18^7.3^3+3^{15}.2^{15}}{2^{10}.6^{15}+3^{14}.15.4^{13}}\) - Hoc24

\(=\dfrac{2^{18}\cdot3^3\cdot\left(3^2\cdot2\right)^7+3^{15}\cdot2^{15}}{2^{10}\cdot2^{15}\cdot3^{15}+3^{14}\cdot3\cdot5\cdot2^6}\)

\(=\dfrac{2^{25}\cdot3^{17}+3^{15}\cdot2^{15}}{2^{25}\cdot3^{15}+3^{15}\cdot5\cdot2^6}\)

\(=\dfrac{2^{15}\cdot3^{15}\left(2^{10}\cdot3^2+1\right)}{2^6\cdot3^{15}\left(2^{19}+5\cdot1\right)}=\dfrac{2^9\cdot9217}{524293}\)

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

\(\frac{3^{17}\cdot81^{11}}{27^{10}\cdot9^{15}}\)

\(=\frac{3^{17}\cdot\left(3^4\right)^{11}}{\left(3^3\right)^{10}\cdot\left(3^2\right)^{15}}\)

\(=\frac{3^{17}\cdot3^{44}}{3^{30}\cdot3^{30}}\)

\(=\frac{3^{61}}{3^{60}}\)

\(=3\)

\(\frac{9^2\cdot2^{11}}{16^2\cdot6^3}\)

\(=\frac{\left(3^2\right)^2\cdot2^{11}}{\left(2^4\right)^2\cdot\left(2\cdot3\right)^3}\)

\(=\frac{3^4\cdot2^{11}}{2^8\cdot2^3\cdot3^3}\)

\(=\frac{3^4\cdot2^{11}}{2^{11}\cdot3^3}\)

\(=\frac{3^4}{3^3}\)

\(=3\)

\(\frac{2^{10}\cdot3^{31}+2^{40}\cdot3^6}{2^{11}\cdot3^{31}+2^{41}\cdot3^6}\)

\(=\frac{2^{10}\cdot3^{31}+2^{40}\cdot3^6}{2\cdot\left(2^{10}\cdot3^{31}+2^{40}\cdot3^6\right)}\)

\(=\frac{1}{2}\)

1/ \(\frac{9.5^{20}.27^9-3.9^{15}.25^9}{7.3^{29}.125^6-3.3^9.15^{19}}\)

\(=\frac{5^{20}.3^{29}-3^{31}.5^{18}}{7.3^{29}.5^{18}-3^{29}.5^{19}}=\frac{3^{29}.5^{18}.\left(25-9\right)}{3^{29}.5^{18}.\left(7-5\right)}=\frac{16}{2}=8\)

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