K
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DK
1 tháng 6 2021
\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{\left(2^4\right)^3.3^{10}+3.5.2^3.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)
\(=\dfrac{2^{12}.3^{10}+3.5.2^3.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}+5.2^{12}.3^{10}}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)\(=\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.7}=\dfrac{2.2}{7}=\dfrac{4}{7}\)
LD
3
AM
20 tháng 6 2015
\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+3^{11}.2^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3+1\right)}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}\left(2.3+1\right)}=\frac{2.6}{3.7}=\frac{12}{21}=\frac{4}{7}\)
MN
1
AH
Akai Haruma
Giáo viên
16 tháng 7
Lời giải:
Gọi biểu thức là $A$.
\(A=\frac{(2^4)^3.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\\ =\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}(2.3+1)}\\ =\frac{2^{12}.3^{10}(1+5)}{7.2^{11}.3^{11}}=\frac{2^{12}.3^{10}.2.3}{7.2^{11}.3^{11}}\\ =\frac{2^{13}.3^{11}}{7.2^{11}.3^{11}}=\frac{2^2}{7}=\frac{4}{7}\)
NK
0
TM
2
6 tháng 2 2016
\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+3^{11}.2^{11}}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}\left(2.3+1\right)}=\frac{2.6}{3.7}=\frac{12}{21}=\frac{4}{7}\)
6 tháng 2 2016
=\(\frac{2^{13}\cdot3^{10}+2^3\cdot3\cdot5\cdot2^9\cdot3^9}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
=\(\frac{2^{12}\cdot3^{10}\cdot\left(1+2\cdot5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3+1\right)}\)
=\(\frac{2\cdot11}{3\cdot7}\)
duyệt nha các bn
=\(\frac{22}{21}\)
TN
0
22 tháng 5 2022
a: \(A=\dfrac{2^{12}\cdot3^{10}+2^3\cdot2^9\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot7}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
b: \(B=\left(\dfrac{12}{105}+\dfrac{9^{15}}{3}\right)\cdot\dfrac{1}{3}\cdot\dfrac{6^8}{6^4\cdot2^4}\)
\(=\dfrac{12+35\cdot9^{15}}{105}\cdot\dfrac{1}{3}\cdot3^4\)
\(=\dfrac{12+35\cdot9^{15}}{105}\cdot3^3=\dfrac{9\left(12+35\cdot9^{15}\right)}{35}\)
DN
0
\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
=\(\dfrac{\left(4^2\right)^3.3^{10}+120.1}{4^6.3^{12}+6^{11-9}}\)
=\(\dfrac{4^{2.3}.1+120}{4^6.3^{12-10}+6^2}\)
=\(\dfrac{4^6+120}{4^6.3^2+6^2}\)
=\(\dfrac{4096+120}{4096.9+36}\)
=\(\dfrac{4216}{36864+36}\)
=\(\dfrac{4216}{36900}=\dfrac{2063}{18450}\)
sai rồi ba