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a)Ta có:\(3^{30}=\left(3^3\right)^{10}=27^{10}\)
\(5^{20}=\left(5^2\right)^{10}=25^{10}\)
Vì \(27^{10}>25^{10}\Rightarrow3^{30}>5^{20}\)
Do 27>25 nên \(27^{10}>25^{10}\)\(hay\) \(3^{30}>5^{20}\)
còn câu b thì mk chưa tính ra
a )( 2/5+2/9-2/11)/(8/5+8/9-8/11)=2*(1/5+1/9-1/11)/8*(1/5+1/9-1/11)=2/8=1/4
\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+3^{11}.2^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3+1\right)}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}\left(2.3+1\right)}=\frac{2.6}{3.7}=\frac{12}{21}=\frac{4}{7}\)
Lời giải:
Gọi biểu thức là $A$.
\(A=\frac{(2^4)^3.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\\ =\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}(2.3+1)}\\ =\frac{2^{12}.3^{10}(1+5)}{7.2^{11}.3^{11}}=\frac{2^{12}.3^{10}.2.3}{7.2^{11}.3^{11}}\\ =\frac{2^{13}.3^{11}}{7.2^{11}.3^{11}}=\frac{2^2}{7}=\frac{4}{7}\)
\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+3^{11}.2^{11}}=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}\left(2.3+1\right)}=\frac{2.6}{3.7}=\frac{12}{21}=\frac{4}{7}\)
=\(\frac{2^{13}\cdot3^{10}+2^3\cdot3\cdot5\cdot2^9\cdot3^9}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
=\(\frac{2^{12}\cdot3^{10}\cdot\left(1+2\cdot5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3+1\right)}\)
=\(\frac{2\cdot11}{3\cdot7}\)
duyệt nha các bn
=\(\frac{22}{21}\)
A=\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.2^9.3^9}{\left(2^2\right)^6.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)
\(=\frac{2.6}{3.7}\)\(=\frac{4}{7}\)
a) 330 và 520
330 = (33)10 = 2710
520 = (52)10 = 2510
=> 2710 > 2510
hay 330 > 520
mình mới học lớp 4 nên ko giải được