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![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{2^{19}\cdot3^9\cdot5+3^{15}\cdot5\cdot2^{18}}{2^{19}\cdot3^9-2^{20}\cdot3^{10}}\)
\(=\dfrac{2^{18}\cdot3^9\cdot5\left(2+3^6\right)}{2^{19}\cdot3^9\left(1-2\cdot3\right)}=\dfrac{1}{2}\cdot\dfrac{5}{-5}\cdot\left(2+3^6\right)\)
\(=\dfrac{-731}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tử số: 2^19 x (3^3)^3 x 5+15 x 4^9 x(3^2)^4
=2^19 x3^9x5 + 15 x(2^2)^9 x 3^8
= 2^19 x 3^9 x 5 +3 x 5 x 2^18 x 3^8
= 2^19 x 3^9 x 5+ 3^9 x 5 x 2^18
= 5 x 3^9 x 2^18 (2+1)
=5 x 3^10 x 2^18
Mẫu số
= (2 x 3)^9 x 2^10 -12^10
= 2^9 x 3^9 x 2^10 - (2^2x3)^10
= 2^9 x 3^9 x 2^10 -2^20 x 3^10
= 2^19 x 3^9 - 2^20 x 3^10
= 2^19 x 3^9 (1-2 x 3)
= 2^19 x 3^9 x(-5)
Chia cả tử và mẫu ta có
(5 x 3^10 x 2^18) / (2^19 x 3^9 x (-5)) = -3/2
\(H=\frac{2^{19}.27^3.5-15.\left(-4\right)^9.9^4}{6^9.2^{10}-\left(-12\right)^{10}}\)
\(\Rightarrow\)\(H=\frac{2^{19}.3^9.5-3.5-1.2^{18}.3^8}{2^9.3^9.2^{10}-6^{10}.2^{10}}=\frac{2^{19}.3^9.5-3^9.5-2^{18}}{2^{19}.3^9-3^{10}.2^{10}.2^{10}}=\frac{2^{19}.3^9.5-3^9.5-2^{18}}{2^{19}.3^9-3^{10}.2^{20}}\)
\(\Rightarrow H=\frac{2^{18}.3^9.5\left(2-1\right)}{2^{19}.3^9.\left(1-3.2\right)}=\frac{5}{2.\left(-5\right)}=\frac{-1}{2}\)
Vậy \(H=\frac{-1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
kết quả cuối cùng =1/2
/ phần nha bạn
nếu đúng thì tick cho mk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
=\(\dfrac{2^{19}.27^3-15.\left(-4\right)^9.9^4}{6^9.2^{10}+\left(-12\right)^{10}}\)
=\(\dfrac{2^{19}.3^9-2^{18}.3^9.5}{6^9.2^{10}+6^{10}.2^{10}}\)
=\(\dfrac{2^{18}.3^9\left(2-2.5\right)}{6^9.2^{10}\left(1+6\right)}\)
=\(\dfrac{2^{18}.3^9.\left(-8\right)}{3^9.2^{19}.7}\)
=\(\dfrac{-8}{14}=\dfrac{-4}{7}\)
bạn làm sai rùi kết quả là \(\dfrac{1}{2}\) mình hỏi thầy rùi
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,\(\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)=\(\dfrac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{6^9.2^{10}+6^{10}.2^{10}}\)
=\(\dfrac{2^{19}.3^9+2^{18}.3^9.5}{6^9.2^{10}.\left(1+6\right)}\)=\(\dfrac{2^{18}.3^9.\left(2+5\right)}{6^9.2^{10}.7}\)=\(\dfrac{2^{18}.3^9}{6^9.2^{10}}=\dfrac{2^{10}.2^8.3^9}{2^9.3^9.2^{10}}=\dfrac{2^8}{2^8.2}=\dfrac{1}{2}\)
b, \(\dfrac{\left(\dfrac{-1}{2}\right)^3-\left(\dfrac{3}{4}\right)^3.\left(-2\right)^2}{2.\left(-1\right)^5+\left(\dfrac{3}{4}\right)^2-\dfrac{3}{8}}=\dfrac{\dfrac{-1}{8}-\dfrac{27}{64}.4}{-2+\dfrac{9}{16}-\dfrac{3}{8}}=\dfrac{\dfrac{-1}{8}-\dfrac{27}{16}}{\dfrac{-23}{16}-\dfrac{3}{8}}=\dfrac{\dfrac{-29}{16}}{\dfrac{-29}{16}}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
C=\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\frac{1}{2}=0,5\)
D=\(\frac{4^7.2^8}{3.2^{15}.16^2-5.2^2.\left(2^{10}\right)^2}\)\(=1\)
Đúng,Đúng,Đúng,...!^-^
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{2^{19}+27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(=\dfrac{2^{19}+\left(3^3\right)^3+5.3.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(3.4\right)^{10}}\)
\(=\dfrac{2^{19}.3^9+3.5.2^{18}.3^8}{3^9.2^9.2^{10}+3^{10}.4^{10}}\)
\(=\dfrac{2^{19}.3^9+5.2^{18}.3^9}{3^9.2^{19}+3^{10}.\left(2^2\right)^{10}}\)
\(=\dfrac{2^{18}.3^9.\left(2.5\right)}{3^9.2^{19}+3^{10}.2^{20}}\)
\(=\dfrac{2^{18}.3^9.7}{2^{19}.3^9.\left(1+3.2\right)}\)
\(=\dfrac{7}{2\left(1+6\right)}\)
\(=\dfrac{7}{2.7}\)
\(=\dfrac{1}{2}\)
a) \(5^{20}và2550^{10}\)
\(5^{20}=\left(5^2\right)^{10}=25^{10}< 2550^{10}\)
=> \(5^{20}< 2550^{10}\)
b) \(999^{10}và999999^5\)
\(999^{10}=\left(999^2\right)^5=1998^5< 999999^5\)
=> \(999^{10}< 999999^5\)
c) \(\left(\dfrac{-1^{300}}{5}\right)và\left(\dfrac{-1^{500}}{3}\right)\)
\(\left(\dfrac{-1^{300}}{5}\right)=\dfrac{-1}{5}\)
\(\left(\dfrac{-1^{500}}{3}\right)=\dfrac{-1}{3}\)
\(\dfrac{-1}{5}=\dfrac{-3}{15}\)
\(\dfrac{-1}{3}=\dfrac{-5}{15}\)
=> \(\dfrac{-3}{15}>\dfrac{-5}{15}\)
=> \(\left(\dfrac{-1^{300}}{5}\right)>\left(\dfrac{-1^{500}}{3}\right)\)
\(H=\dfrac{2^{19}\cdot3^9\cdot5+3\cdot5\cdot2^{18}\cdot3^8}{2^9\cdot2^{10}\cdot3^9-2^{20}\cdot3^{10}}\)
\(=\dfrac{2^{19}\cdot3^9\cdot5+2^{18}\cdot3^9\cdot5}{2^{19}\cdot3^9-2^{20}\cdot3^{10}}\)
\(=\dfrac{2^{18}\cdot3^9\cdot5\cdot\left(2+1\right)}{2^{19}\cdot3^9\cdot\left(1-6\right)}\)
\(=\dfrac{1}{2}\cdot\dfrac{5\cdot3}{-5}=-\dfrac{1}{2}\cdot3=-\dfrac{3}{2}\)