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a) \(\frac{10^4.81-16.15^2}{\left(-8\right)^4.3^{12}+6^{11}}=\frac{10^4.3^4-4^2.15^2}{8^4.3^{12}+6^{11}}=\frac{30^4-60^2}{2^{12}.3^{12}+6^{11}}=\frac{\left(30^2\right)^2-60^2}{6^{12}+6^{11}}\)
\(=\frac{900^2-60^2}{6^{11}.\left(6+1\right)}=\frac{60^2.\left(15^2-1\right)}{6^{11}.7}=\frac{60^2.224}{6^{11}.7}=\frac{2^9.3^2.5^2.7}{2^{11}.3^{11}.7}=\frac{5^2}{2^2.3^9}=\frac{25}{78732}\)
a) 227 = (23)9 = 89
318 = (32)9 = 99
Do: 8 < 9 nên 89 < 99 hay 227 < 318
b) 291 = (213)7 = 81927
535 = (55)7 = 31257
Do: 8192 > 3125 nên 291 > 535
c) 2225 = (23)75 = 875
3150 = (32)75 = 975
Do: 8 < 9 nên 2225 < 3150
d) 912 = (32)12 = 324
277 = (33)7 = 321
Do: 24 > 21 nên 912 > 277
So sánh: \(\left(-\dfrac{1}{6}\right)^{100}\) và \(\left(-\dfrac{1}{2}\right)^{500}\)
Ta có: \(\left(-\dfrac{1}{2}\right)^{500}=\left[\left(-\dfrac{1}{2}\right)^5\right]^{100}=\left(-\dfrac{1}{32}\right)^{100}=\left(\dfrac{1}{32}\right)^{100}\)
\(\left(-\dfrac{1}{6}\right)^{100}=\left(\dfrac{1}{6}\right)^{100}\)
Vì: \(\left(\dfrac{1}{32}\right)^{100}< \left(\dfrac{1}{6}\right)^{100}\Rightarrow\left(-\dfrac{1}{6}\right)^{100}>\left(-\dfrac{1}{2}\right)^{500}\)
\(\left(-\dfrac{1}{6}\right)^{100}\)
\(\left(-\dfrac{1}{2}\right)^{500}\)\(=\left(\left(-\dfrac{1}{2}\right)^5\right)^{100}=\left(\dfrac{-1}{2^5}\right)^{100}=\left(-\dfrac{1}{32}\right)^{100}\)
Vì \(-\dfrac{1}{6}< \left(-\dfrac{1}{32}\right)\)
=>\(\left(-\dfrac{1}{6}\right)^{100}< \left(-\dfrac{1}{32}\right)^{100}\)
Vậy \(\left(-\dfrac{1}{6}\right)^{100}< \left(-\dfrac{1}{2}\right)^{500}\)
#H ( ~~~~~~~Study well~~~~~~~~)
a)\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\Leftrightarrow\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}\)
Áp dụng tc dãy tỉ
\(\frac{x}{\frac{3}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{5}{4}}=\frac{x+y+z}{\frac{3}{2}+\frac{4}{3}+\frac{5}{4}}=\frac{49}{\frac{49}{12}}=12\)
Với \(\frac{x}{\frac{3}{2}}=12\Rightarrow x=18\)
Với \(\frac{y}{\frac{4}{3}}=12\Rightarrow y=16\)
Với \(\frac{z}{\frac{5}{4}}=12\Rightarrow z=15\)
b)\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\Leftrightarrow\frac{a^2}{4}=\frac{b^2}{9}=\frac{2c^2}{32}\)
Áp dụng tc dãy tỉ
\(\frac{a^2}{4}=\frac{b^2}{9}=\frac{2c^2}{32}=\frac{a^2-b^2+2c^2}{4-9+32}=\frac{108}{27}=4\)
Với \(\frac{a^2}{4}=4\Rightarrow a=4\)
Với \(\frac{b^2}{9}=4\Rightarrow b=6\)
Với \(\frac{2c^2}{32}=4\Rightarrow c=8\)
nhanhh mn oiw