Question

Помогите пожалуйста, пока решал получил 3 инсульта

Answer 1

$( \frac{1}{7} ) {}^{ - 8} \times ( \frac{1}{7} ) {}^{7} = ( \frac{1}{7} ) {}^{ - 8 + 7} = ( \frac{1}{7} ) {}^{ - 1} = 7$

$(( \frac{1}{3} ) {}^{ - 1} ) {}^{ - 4} = ( \frac{1}{3} ) {}^{ - 1 \times ( - 4)} = ( \frac{1}{3} ) {}^{4} = \frac{1}{81}$

$81 \times 3 {}^{ - 5} = {3}^{4} \times 3 {}^{ - 5} = 3 {}^{4 - 5} = {3}^{ - 1} = \frac{1}{3}$

${32}^{ - 2} \times {4}^{4} = ( {2}^{5} ) {}^{ - 2} \times ( {2}^{2} ) {}^{4} = {2}^{ 5\times ( - 2) + 2 \times 4} = \\ = 2 {}^{ - 10 + 8} = {2}^{ - 2} = \frac{1}{ {2}^{2} } = \frac{1}{4} = 0.25$

$\frac{ {81}^{ - 5} \times {9}^{ - 8} }{ {27}^{ - 11} } = \frac{( {3}^{4}) {}^{ - 5} \times ( {3}^{2} ) {}^{ - 8} }{( {3}^{3}) {}^{ - 11} } = 3 {}^{4 \times ( - 5) + 2 \times ( - 8) - 3 \times ( - 11)} = \\ = {3}^{ - 20 - 16 + 33} = {3}^{ - 36 + 33} = {3}^{ - 3} = \frac{1}{ {3}^{3} } = \frac{1}{27}$