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\(\frac{3^{10}+6^2}{5\cdot3^8+20}\)
\(=\frac{3^{10}+\left(3\cdot2\right)^2}{5\left(3^8+4\right)}\)
\(=\frac{3^{10}+3^2\cdot2^2}{5\left(3^8+4\right)}\)
\(=\frac{3^2\left(3^8+4\right)}{5\left(3^8+4\right)}\)
\(=\frac{9}{5}\)
So sánh: 7215 và 321.969
7215=(23.32)15=(23)15.(32)15=245.330
321.969=321.(3.25)9=321.39.(25)9=321+9.245=330.245
Vì 245.330 = 3030.245
Nên 7215 = 321.969
Bây giờ mình sẽ viết đầy đủ hơn nhé:
a) \(\frac{28^{15}.3^{17}}{84^{16}}=\frac{\left(28.3\right)^{15}.3^2}{\left(28.3\right)^{15}.84}=\frac{9}{84}=\frac{3}{28}\)
b)\(\frac{3^{10}+6^2}{5.3^8+20}=\frac{3^{10}+2^2.3^2}{5.3^8+20}=\frac{3^2.\left(3^8+2^2\right)}{5.\left(3^8+2^2\right)}\)\(=\frac{9}{5}\)
a) \(\frac{28^{15}.3^{17}}{84^{16}}=\frac{\left(2^2.7\right)^{15}.3^{17}}{\left(2^2.3.7\right)^{16}}=\frac{2^{30}.7^{15}.3^{17}}{2^{32}.3^{16}.7^{16}}=\frac{3}{2^2.7}=\frac{3}{28}\)
b) \(\frac{3^{10}+6^2}{5.3^8+20}=\frac{3^{10}+\left(2.3\right)^2}{5.3^9+2^2.5}=\frac{3^{10}+2^2.3^2}{5\left(3^8.2^2\right)}=\frac{3^2.\left(3^8+2^2\right)}{5.\left(3^8+2^2\right)}=\frac{3^2}{5}=\frac{9}{5}\)
\(P=\dfrac{4\cdot36^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{4\cdot\left(2^2\cdot3^2\right)^4-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
1:a)\(\frac{28^{15}\cdot3^{17}}{84^{16}}\)=\(\frac{28^{15}\cdot3^{15}\cdot3^2}{84^{16}}\)=\(\frac{\left(28^{15}\cdot3^{15}\right)\cdot3^2}{84^{16}}\)=\(\frac{84^{15}\cdot9}{84^{16}}\)=\(\frac{9}{84}\)=\(\frac{3}{28}\)
b)\(\frac{3^{10}+6^2}{5\cdot3^8+20}\)=\(\frac{3^2\cdot3^8+2^2\cdot3^2}{5\cdot3^8+5\cdot4}\)=\(\frac{9\cdot3^8+4\cdot9}{5\cdot\left(3^8+4\right)}\)=\(\frac{9\cdot\left(3^8+4\right)}{5\cdot\left(3^8+4\right)}\)=\(\frac{9}{5}\)
Xét vế trái ta có:72^15=3^15*24^15=3^15*24^9*24^6
=3^15*24^9*12^6*2^6=3^15*24^9*12^6*4^3(1)
Xét vế phải ta có: 3^21*96^9=3^15*3^6*24^9*4^9
=3^15*3^6*24^9*4^6*4^3=3^15*24^9*(3^6*4^6*4^3)
=3^15*24^9*(12^6*4^3)(2)
từ (1) và (2)=>72^15=3^21*96^9
\(\frac{3^{10}+6^2}{5.3^8+20}\)\(=\frac{3^2+2^2.3^2}{5+2^2.5}=\frac{3^2\left(1+2^2\right)}{5\left(1+2^2\right)}\)\(=\frac{3^2}{5}\)\(=\frac{9}{5}\)