Question

решите пожалуйста срочно!!!!!!!​

Answer 1

▪1.

$a) \: \: \: {(3a + b)}^{2} = 9 {a}^{2} + 6ab + {b}^{2} \\ \\ b) \: \: \: (y - \frac{1}{5}x )(y + \frac{1}{5}x ) = {y}^{2} - \frac{1}{25} {x}^{2}$

▪2.

$a) \: \: \: {m}^{2} - 81 {n}^{2} = (m - 9n)(m + 9n) \\ \\ б) \: \: \: 25 + 10p + {p}^{2} = {(5 + p)}^{2} \\ \\ в) \: \: \: 6 {b }^{7} - 24 \: {b} ^{4} = 6 {b}^{4} ( {b}^{3} - 4)$

▪3.

$( {17.6}^{2} - {2.4}^{2} ) \div 10 = (17.6 - 2.4)(17.6 + 2.4) \div 10 = 15.2 \times 20 \div 10 = 15.2 \times 2 = 30.4$

▪4.

$y(7x - y) + {(x - y)}^{2} = 7xy - {y}^{2} + {x}^{2} - 2xy + {y}^{2} = 5xy + {x}^{2} = x(5y + x) \\ \\ - 1 \times (5 \times 0.2 + ( - 1)) = - (1 - 1) = 0$

▪5.

$a) \: \: \: 20 {m}^{2} - 5 {n}^{4} = 5(4 {m}^{2} - {( {n}^{2}) }^{2} ) = 5(2m - {n}^{2} )(2m + {n}^{2} ) \\ \\ b) \: \: \: 64 {c}^{3} - {b}^{3} = (4c - b)(16 {c}^{2} + 4bc + {b}^{2} )$

▪6.

${(x + 1)}^{2} = 36 \\ {x}^{2} + 2x + 1 - 36 = 0 \\ {x}^{2} + 2x - 35 = 0 \\ d = 4 - 4 \times ( - 35) = 4 + 140 = 144 \\ x_{1} = \frac{ - 2 + \sqrt{144} }{2} = \frac{ - 2 + 12}{2} = \frac{10}{2} = 5 \\ x_{2} = \frac{ - 2 - \sqrt{144} }{2} = \frac{ - 2 - 12}{2} = \frac{ - 14}{2} = - 7$