Question

помогите пожалуйста алгебра 9 класс

Answer 1

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:

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

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

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

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