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C = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right).......\left(\frac{1}{9^2}-1\right)\left(\frac{1}{10^2}-1\right)\)
= \(\frac{-3}{2^2}.\frac{-8}{3^2}.............\frac{-80}{9^2}.\frac{-99}{10^2}\)
= \(-\left(\frac{1.3}{2^2}.\frac{2.4}{3^2}..........\frac{8.10}{9^2}.\frac{9.11}{10^2}\right)\)
= \(-\frac{\left(1.2.3......8.9\right)\left(3.4.5........10.11\right)}{\left(2.3.4......9.10\right)\left(2.3.4......9.10\right)}=-\frac{1.11}{10.2}=-\frac{11}{20}\)
\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)
\(\left(\frac{27}{64}\right)^{15}=\frac{\left(3^3\right)^{15}}{\left(2^6\right)^{15}}=\frac{3^{45}}{2^{90}}=\left(\frac{3}{2^2}\right)^{45}\)
\(\left(\frac{81}{256}\right)^{10}=\frac{\left(3^4\right)^{10}}{\left(2^8\right)^{10}}=\frac{3^{40}}{2^{80}}=\left(\frac{3}{2^2}\right)^{40}\)
Do \(\left(\frac{3}{2^2}\right)^{45}<\left(\frac{3}{2^2}\right)^{40}\Rightarrow\left(\frac{27}{64}\right)^{15}<\left(\frac{81}{256}\right)^{10}\)
a, \(\frac{16}{2^n}=2\Leftrightarrow2\cdot2^n=16\Leftrightarrow2^{n+1}=2^4\Leftrightarrow n+1=4\Rightarrow n=3\)
b,\(\frac{\left(-3\right)^n}{81}=-27\Leftrightarrow\left(-3\right)^n=-3^3.3^4\Leftrightarrow\left(-3\right)^n=\left(-3\right)^7\Rightarrow n=7\)
c,\(8^n:2^n=4\Leftrightarrow\left(2^3\right)^n:2^n=2^2\Leftrightarrow2^{3n}:2^{2n}=2^2\Leftrightarrow2^{3n-2n}=2^2\Leftrightarrow2^n=2^2\Leftrightarrow n=2\)
câu 1. \(\frac{16}{2^n}\)=2 =>2n=16x2=32
2n=25 => n=5
câu 2.\(\frac{\left(-3\right)^n}{81}\)=-27=33 =>(-3)n =81x33
=(-3)4x(-3)3
=(-3)7
câu 3. 8n:4n=4=>(8:4)n=4
2n=4 =22
=>n=2
a)\(\left(\frac{-1}{3}\right)^3\cdot x=\frac{1}{81}\) \(< =>\frac{-1}{27}x=\frac{1}{81}\)\(< =>x=\frac{-1}{3}\)
\(\left(\frac{1}{27}\right)^{23}=\frac{1^{23}}{27^{23}}=\frac{1}{\left(3^3\right)^{23}}=\frac{1}{3^{69}}\)
\(\left(\frac{1}{81}\right)^{16}=\frac{1^{16}}{81^{16}}=\frac{1}{\left(3^4\right)^{16}}=\frac{1}{3^{64}}\)
Vì 369 > 364
\(\frac{1}{3^{69}}< \frac{1}{3^{64}}\)
\(\left(\frac{1}{27}\right)^{23}=\frac{1^{23}}{27^{23}}=\frac{1}{\left(3^3\right)^{23}}=\frac{1}{3^{69}}\)
\(\left(\frac{1}{81}\right)^{16}=\frac{1^{16}}{81^{16}}=\frac{1}{\left(3^4\right)^{16}}=\frac{1}{3^{64}}\)
Vì 369 > 364
=> \(\frac{1}{3^{69}}< \frac{1}{3^{64}}\)
=> \(\left(\frac{1}{27}\right)^{23}< \left(\frac{1}{81}\right)^{16}\)