Задание 575

Решение:

• \({\largeа)}\ \vphantom{\frac{0}{0}}10^4=10\cdot10\cdot10\cdot10=10\ 000\)
• \(\phantom{\largeа)}\ \vphantom{\frac{0}{0}}10^3=10\cdot10\cdot10=1000\)
• \(\phantom{\largeа)}\ \vphantom{\frac{0}{0}}10^2=10\cdot10=100\)
• \(\phantom{\largeа)}\ \vphantom{\frac{0}{0}}10^1=10\)
• \(\phantom{\largeа)}\ \vphantom{\frac{0}{0}}10^0=1\)
• \(\phantom{\largeа)}\ 10^{{-}1}=\frac{1}{10^1}=\frac{1}{10}\)
• \(\phantom{\largeа)}\ 10^{{-}2}=\frac{1}{10^2}=\frac{1}{10\cdot10}=\frac{1}{100}\)
• \(\phantom{\largeа)}\ 10^{{-}3}=\frac{1}{10^3}=\frac{1}{10\cdot10\cdot10}=\frac{1}{1000}\)
• \(\phantom{\largeа)}\ 10^{{-}4}=\frac{1}{10^4}=\frac{1}{10\cdot10\cdot10\cdot10}=\frac{1}{10\ 000}\)

• \({\largeб)}\ \vphantom{\frac{0}{0}}2^5=2\cdot2\cdot2\cdot2\cdot2=32\)
• \(\phantom{\largeб)}\ \vphantom{\frac{0}{0}}2^4=2\cdot2\cdot2\cdot2=16\)
• \(\phantom{\largeб)}\ \vphantom{\frac{0}{0}}2^3=2\cdot2\cdot2=8\)
• \(\phantom{\largeб)}\ \vphantom{\frac{0}{0}}2^2=2\cdot2=4\)
• \(\phantom{\largeб)}\ \vphantom{\frac{0}{0}}2^1=2\)
• \(\phantom{\largeб)}\ \vphantom{\frac{0}{0}}2^0=1\)
• \(\phantom{\largeб)}\ 2^{{-}1}=\frac{1}{2^1}=\frac{1}{2}\)
• \(\phantom{\largeб)}\ 2^{{-}2}=\frac{1}{2^2}=\frac{1}{2\cdot2}=\frac{1}{4}\)
• \(\phantom{\largeб)}\ 2^{{-}3}=\frac{1}{2^3}=\frac{1}{2\cdot2\cdot2}=\frac{1}{8}\)
• \(\phantom{\largeб)}\ 2^{{-}4}=\frac{1}{2^4}=\frac{1}{2\cdot2\cdot2\cdot2}=\frac{1}{16}\)
• \(\phantom{\largeб)}\ 2^{{-}5}=\frac{1}{2^5}=\frac{1}{2\cdot2\cdot2\cdot2\cdot2}=\frac{1}{32}\)

• \({\largeв)}\ \vphantom{\frac{0}{0}}({-}3)^3=({-}3)\cdot({-}3)\cdot({-}3)={-}27\)
• \(\phantom{\largeв)}\ \vphantom{\frac{0}{0}}({-}3)^2=({-}3)\cdot({-}3)=9\)
• \(\phantom{\largeв)}\ \vphantom{\frac{0}{0}}({-}3)^1={-}3\)
• \(\phantom{\largeв)}\ \vphantom{\frac{0}{0}}({-}3)^0=1\)
• \(\phantom{\largeв)}\ ({-}3)^{{-}1}=\frac{1}{ ({-}3)^1}=\frac{1}{{-}3}={-}\frac{1}{3}\)
• \(\phantom{\largeв)}\ ({-}3)^{{-}2}=\frac{1}{ ({-}3)^2}=\frac{1}{ ({-}3)\cdot({-}3)}=\frac{1}{9}\)
• \(\phantom{\largeв)}\ ({-}3)^{{-}3}=\frac{1}{ ({-}3)^3}=\frac{1}{ ({-}3)\cdot({-}3)\cdot({-}3)}=\frac{1}{{-}27}={-}\frac{1}{27}\)