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a) \(2^{-1}\cdot2^n+4\cdot2^n=9\cdot2^5\)
\(\Rightarrow2^n\cdot\left(2^{-1}+4\right)=9\cdot2^5\)
\(\Rightarrow2^n\cdot4,5=288\)
\(\Rightarrow2^n=64\)
\(\Rightarrow n=6\)
b) \(2^m-2^n=1984\)
\(\Rightarrow2^n\cdot\left(2^{m-n}-1\right)=2^6\cdot31\)
\(\Rightarrow\left\{{}\begin{matrix}2^n=2^6\\2^{m-n}-1=31\end{matrix}\right.\)
\(\Rightarrow n=6\)
\(\Rightarrow2^{m-n}=32\Rightarrow m-n=5\Rightarrow m=11\)
a. \(\frac{20^5.5^{10}}{100^5}\)
\(=\frac{20^5.\left(5^2\right)^5}{100^5}\)
\(=\frac{20^5.25^5}{100^5}\)
\(=\frac{500^5}{100^5}\)
\(=\left(\frac{500}{100}\right)^5\)
\(=5^5=3125\)
b. \(\frac{\left(0,9\right)^5}{\left(0,3\right)^6}\)
\(=\frac{\left(0,9\right)^5}{\left(0,3\right)^5.0,3}\)
\(=\left(\frac{0,9}{0,3}\right)^5.\frac{1}{0,3}\)
\(=3^5.\frac{1}{0,3}\)
\(=810\)
c. \(\frac{6^3+3.6^2+3^3}{-13}\)
\(=\frac{\left(3.2\right)^3+3.\left(3.2\right)^2+3^3}{-13}\)
\(=\frac{3^3\left(2^3+2^2+1\right)}{-13}\)
\(=\frac{3^3.13}{-13}\)
\(=\left(-3\right)^3\)
\(=-27\)
Bài 1:
Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Leftrightarrow2x=\frac{1440}{144}=10\)
\(\Rightarrow x=5\)
Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
a) \(2^5+8\left[\left(-2\right)^3:\frac{1}{2}\right]^0-\left(\frac{1}{2}\right)^3\times2+\left(-2\right)^3\)
\(=32+8\times1-\frac{1}{8}\times2+\left(-8\right)\)
\(=32+8-\frac{1}{4}+\left(-8\right)\)
\(=40-\frac{1}{4}+\left(-8\right)\)
\(=39\frac{3}{4}+\left(-8\right)\)
\(=31\frac{3}{4}\)
b vaf c mai minhf lamf, ht