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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)\(\frac{3^{10}+6^2}{5.3^8+20}=\frac{3^{10}+2^2.3^2}{5.3^8+20}=\frac{3^2.\left(3^8+2^2\right)}{5.\left(3^8+2^2\right)}\)\(=\frac{9}{5}\)
Rut gon bieu thuc ( 2x like )
a)28^15 . 3^17
_____________
84^16
b) 3^10 + 6^2
________________
5.3^8 + 20
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
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{18}{19}.\frac{19}{20}\)
\(A=\frac{1}{20}\)
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)........\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)
\(\Leftrightarrow A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...........\frac{18}{19}.\frac{19}{20}\)
\(\Leftrightarrow A=\frac{1}{20}>\frac{1}{21}\)
\(\Leftrightarrow A>\frac{1}{21}\)
\(B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)................\left(1-\frac{1}{100}\right)\)
\(\Leftrightarrow B=\frac{3}{4}.\frac{8}{9}..................\frac{99}{100}\)
\(B=\frac{1.3}{2^2}.\frac{2.4}{3^2}................\frac{9.11}{50^2}\)
\(B=\frac{11}{50}< \frac{11}{21}\)
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) \(\frac{3^{10}+6^2}{5.3^8+20}=\frac{3^{10}+\left(2.3\right)^2}{5.3^9+2^2.5}=\frac{3^{10}+2^2.3^2}{5\left(3^8.2^2\right)}=\frac{3^2.\left(3^8+2^2\right)}{5.\left(3^8+2^2\right)}=\frac{3^2}{5}=\frac{9}{5}\)