Предмет: Алгебра, автор: werager512

помогите пожалуйста алгебра 9 класс

Приложения:

Ответы

Автор ответа: bbbapho
0

Найдите значение выражения:

1) {7}^{5}  \times  {7}^{ - 7}  =  {7}^{5 - 7}  =  {7}^{ - 2}  =  \frac{1}{ {7}^{2} }  =  \frac{1}{49}

2) {10}^{ - 12}  \times  {10}^{15}  =  {10}^{ - 12 + 15}  =  {10}^{3}  = 1000

3) {5}^{ - 12}  \div  {5}^{ - 16}  =  {5}^{ - 12 + 16}  =  {5}^{4}  = 625

4) {3}^{ - 14}  \times  {3}^{ - 19}  \div  {3}^{ - 34}  =  {3}^{ - 14 - 19 + 34}  =  {3}^{1}  = 3

5) {( {13}^{ - 9}) }^{4}  \times  {( {13}^{ - 2}) }^{ - 18}  =  {13}^{ - 9 \times 4}  \times  {13}^{ - 2 \times ( - 18)}  =  {13}^{ - 36}  \times  {13}^{36}  =  {13}^{0}  = 1

\frac{ {2}^{ - 4} \times  {( {2}^{ - 3}) }^{5}  }{ {( {2}^{ - 8} )}^{2}  \times  {2}^{ - 3} }  =  \frac{ {2}^{ - 4} \times  {2}^{ - 3 \times 5}  }{ {2}^{ - 8 \times 2} \times  {2}^{ - 3}  }  =  \frac{ {2}^{ - 4} \times  {2}^{ - 15}  }{ {2}^{ - 16} \times  {2}^{ - 3}  }  =  \frac{ {2}^{ - 19} }{ {2}^{ - 19} }  =  {2}^{0} = 1

Найдите значение выражения:

 {27}^{ - 3}  \div  {81}^{ - 2}  =  \frac{ {27}^{ - 3} }{( {3}^{ - 2}  \times  {27}^{ - 2}) }  =  \frac{ {27}^{ - 1} }{ {3}^{ - 2} }  =  \frac{ {3}^{ - 1} \times  {9}^{ - 1}  }{ {3}^{ - 2} }  =  {3}^{1}  \times  {9}^{ - 1}  =  \frac{3}{9}  =  \frac{1}{3}

 \frac{ {( - 36)}^{ - 3} \times  {6}^{4}  }{ {216}^{ - 4}  \times  {( - 6)}^{9} }  =  \frac{ {( - 36)}^{ - 3} \times  {6}^{4}  }{ {36}^{ - 4} \times  {6}^{ - 4}  \times  {( - 6)}^{9}  }  =  \frac{ {( - 1)}^{ - 3}  \times  {36}^{ - 3} \times  {6}^{4}  }{ {36}^{ - 4}  \times  {6}^{ - 4}  \times  {( - 1)}^{9} \times  {6}^{9}  }  =  {( - 1)}^{ - 3 - 9}  \times  {36}^{ - 3 + 4}  \times  {6}^{4 + 4 - 9}  =  {( - 1)}^{ - 12}  \times  {36}^{1}  \times  {6}^{ - 1}  =  \frac{36}{6}  = 6

 \frac{ {21}^{5} \times  {3}^{ - 7}  }{ {63}^{ - 2} \times  {7}^{8}  }  =  \frac{ {3}^{5} \times  {7}^{5}  \times  {3}^{ - 7}  }{ {3}^{ - 2} \times  {3}^{ - 2}   \times  {7}^{ - 2}  \times  {7}^{8} }  =  {3}^{5 - 7 + 2 + 2}  \times  {7}^{5 + 2 - 8}  =  {3}^{2}  \times  {7}^{ - 1}  =  \frac{9}{7}

\frac{ {(0.2)}^{ - 6} \times  {25}^{ - 7}  }{ {125}^{ - 3} }   \\ \\  {(0.2)}^{ - 6}  =   {( \frac{1}{5} )}^{ - 6}  =  {5}^{6}  \\  \\  =  \frac{ {5}^{6}  \times  {5}^{ - 7}  \times  {5}^{ - 7} }{ {5}^{ -  3} \times  {5}^{ - 3}  \times  {5}^{ - 3}  }  =  {5}^{6 - 7 - 7 + 3 + 3 + 3}  =  {5}^{1}  = 5

Упростите выражение:

 \frac{1}{3}  {p}^{ - 2}  {q}^{ - 5}  \times  \frac{9}{5}  {p}^{6}  {q}^{3}  =  \frac{3}{5}  {p}^{ - 2 + 6}  \times  {q}^{ - 5 + 3}  =  \frac{3}{5}  {p}^{4}  {q}^{ - 2}  =  \frac{3 {p}^{4} }{5 {p}^{2} }

 - 0.4 {b}^{ - 3}  \times  {c}^{7}  \times 1.5 {b}^{2}  {c}^{ - 6}  =  -  \frac{4}{10}  {b}^{ - 3}  \times  {c}^{7}  \times  \frac{15}{10}  {b}^{2}  {c}^{ - 6}  =  -  \frac{3}{5}  {b}^{ - 3 + 2}  \times  {c}^{7 - 6}  =  -  \frac{3c}{5b}

<...>

5 {a}^{ - 6}  \times  {( - 3 {a}^{ - 2}  {b}^{3} )}^{2}  = 5 {a}^{ - 6}  \times ( -  {3}^{ - 2}    {a}^{4}  {b}^{ - 6} ) =  - 15 {a}^{ - 6 + 4}  {b}^{ - 6}  =  - 15 {a}^{ - 2}  {b}^{ - 6}  =  -  \frac{15}{ {a}^{2} {b}^{6}  }

Выполните действия и приведите полученное выражение к виду, не содержащему степени с отрицательным показателем:

 \frac{17 {x}^{ - 8} }{14 {y}^{ - 12} }   \times  \frac{28 {y}^{2} }{51 {x}^{ - 21} }  =  \frac{2}{3}  \times  {x}^{ - 8 + 21}  \times  {y}^{1 + 12}  =  \frac{2}{3}  {x}^{13}  {y}^{13}

13>0

 - 1.6 {m}^{ - 4}  {n}^{3}   \times  { ( - 2 {m}^{ - 3}  {p}^{ - 6} )}^{3}  =  -  \frac{16}{10}  {m}^{ - 4}  {n}^{3} \times ( {( - 2)}^{ - 3}  {m}^{9}  {p}^{18} ) =  -  \frac{16}{10}  {m}^{ - 4 + 9}  {n}^{3}  {p}^{18} ( -  \frac{1}{8} ) =  \frac{1}{5}  {m}^{5}  {n}^{3}  {p}^{18}

5>0, 3>0, 18>0

2 \frac{1}{4}  {a}^{ - 5} b {(1 \frac{1}{2}  {a}^{ - 1}  {b}^{ - 3} )}^{ - 3}  =  \frac{9}{4}  {a}^{ - 5} b \times  {( \frac{3}{2} )}^{ - 3}  \times  {a}^{3}  {b}^{9}  =  \frac{9 \times 8}{4 \times 27}  {a}^{ - 5 + 3}  {b}^{1 + 9}  =  \frac{4}{3}  {a}^{ - 2}  {b}^{10}  =  \frac{4 {b}^{10} }{3 {a}^{2} }

10>0, 2>0

 {( - 10 {a}^{ - 2} b {c}^{ - 11} )}^{ - 2}  \times  {(0.1b {c}^{ - 2} )}^{ - 3}  =  {( - 10)}^{ - 2}  {a}^{4}  {b}^{ - 2}  {c}^{22}  {(0.1)}^{ - 3}  {b}^{ - 3}  {c}^{6}  =  \frac{1}{100}  \times  \frac{1000}{1}  \times  {a}^{4}  {b}^{ - 5}  {c}^{28}  =  \frac{10 {a}^{4} {c}^{28}  }{ {b}^{5} }

4>0, 28>0, 5>0

 {( -  \frac{1}{5}  {a}^{ - 3} {b}^{ - 7}  )}^{ - 3}  \times  {( - 5 {a}^{2}  {b}^{6} )}^{ - 2}  =  {( -  \frac{1}{5} )}^{ - 3}  {a}^{9}  {b}^{21}  \times   {( - 5)}^{ - 2}  {a}^{ - 4}  {b}^{ - 12}  =  {( - 5)}^{3}  {a}^{9 - 4}   {b}^{21 - 12} \times   {( - 5)}^{ - 2}  =  - 5 {a}^{5}  {b}^{9}

5>0, 9>0

Похожие вопросы