\[\mathbf{Страница\ 54.}\]
\[\boxed{\mathbf{6.}}\]
\[a \cdot b\]
\[6;\ - 28;\ - 40;\ - 80;\]
\[14;\ \frac{2}{5};\ \frac{7}{100};\ \frac{3}{5};\ \ \]
\[1;\ \ 128;\ - 132;120.\]
\[\boxed{\mathbf{7.}}\]
\[\textbf{а)} - 6 \cdot 2 = - 12\]
\[- 4 \cdot ( - 5) = 20\]
\[7 \cdot ( - 3) = - 21\]
\[\textbf{б)} - 7 \cdot 1 = - 7\]
\[- 5 \cdot ( - 1) = 5\]
\[6 \cdot ( - 1) = - 6\]
\[\textbf{в)} - 5 \cdot 4 = - 20\]
\[0 \cdot ( - 10) = 0\]
\[- 9 \cdot ( - 3) = 27\]
\[\boxed{\mathbf{8.}}\]
\[\textbf{а)}\ ( - 2)^{3} = - 8\]
\[( - 5)^{2} = 25\]
\[( - 1)^{2} = 1\]
\[\left( 2\frac{3}{4} \right)^{2} = \frac{11}{4} \cdot \frac{11}{4} = \frac{121}{16} = 7\frac{9}{16}\]
\[\left( \frac{5}{7} \right)^{2} = \frac{25}{49}\]
\[\textbf{б)}\ \left( \frac{3}{10} \right)^{3} = \frac{27}{1000}\]
\[( - 9)^{2} = 81\]
\[( - 7)^{3} = - 343\]
\[\left( 4\frac{2}{3} \right)^{2} = \frac{14}{3} \cdot \frac{14}{3} = \frac{196}{9} = 21\frac{7}{9}\]
\[\left( \frac{11}{12} \right)^{2} = \frac{121}{144}\]
\[\boxed{\mathbf{9.}}\]
\[\textbf{а)}\ ( - 7) \cdot ( - 2) > 0\]
\[- 9 \cdot 5 < 0\]
\[- 6 \cdot ( - 8) > 0\]
\[- 20 \cdot ( - 20) > 0\]
\[\textbf{б)}\ 7 \cdot ( - 2) < 0\]
\[8 \cdot ( - 9) < 0\]
\[- 1 \cdot ( - 11) > 0\]
\[- 2 \cdot 3 < 0\]
\[\textbf{в)} - 13 \cdot 0 = 0\]
\[- 16 \cdot 1 < 0\]
\[0 \cdot ( - 5) = 0\]
\[- 7 \cdot ( - 6) > 0\]
\[\boxed{\mathbf{10.}}\]
\[y = x^{2} - 2x - 3\]
\[y = ( - 2)^{2} - 2 \cdot ( - 2) - 3 =\]
\[= 4 + 4 - 3 = 5\]
\[y = ( - 1)^{2} - 2 \cdot ( - 1) - 3 =\]
\[= 1 + 2 - 3 = 0\]
\[y = 0 - 2 \cdot 0 - 3 = - 3\]
\[y = 1^{2} - 2 \cdot 1 - 3 =\]
\[= 1 - 2 - 3 = - 4\]
\[y = 2^{2} - 2 \cdot 2 - 3 =\]
\[= 4 - 4 - 3 = - 3\]
\[y = \left( \frac{1}{2} \right)^{2} - 2 \cdot \frac{1}{2} - 3 =\]
\[= \frac{1}{4} - 1 - 3 =\]
\[= - 4 + \frac{1}{4} = - 3\frac{3}{4}\]
\[y = \left( \frac{1}{3} \right)^{2} - 2 \cdot \frac{1}{3} - 3 =\]
\[= \frac{1}{9} - \frac{2}{3} - 3 =\]
\[= - 3\frac{6}{9} + \frac{1}{9} = - 3\frac{5}{9}\]
\[y = \left( \frac{1}{4} \right)^{2} - 2 \cdot \frac{1}{4} - 3 =\]
\[= \frac{1}{16} - \frac{1}{2} - 3 =\]
\[= - 3\frac{8}{16} + \frac{1}{16} = - 3\frac{7}{16}\]
\[y = \left( \frac{1}{5} \right)^{2} - 2 \cdot \frac{1}{5} - 3 =\]
\[= \frac{1}{25} - \frac{2}{5} - 3 =\]
\[= - 3\frac{10}{25} + \frac{1}{25} = - 3\frac{9}{25}\]