Question

пожаоуйста Алгебра 7 класс ​

Answer 1

Объяснение:

1. 1)

$( {2x)}^{2} + 2 \times 2x \times 9 + {9}^{2} = {4x}^{2} + 36x + 81$

2)

${3x}^{3} \times {3x}^{3} + {3x}^{3} \times 4y - 4xy \times {3x}^{3} - 4xy \times 4y = {9x}^{6} + {12x}^{3} y - {12x}^{4} y - {16xy}^{2}$

3)

$( { - 3a)}^{2} - 2 \times ( - 3)a \times 8b + ( {8b)}^{2} = {9a}^{2} + 48ab + {64b}^{2}$

4)

${ - 5m}^{2} \times {5m}^{5} - {5m}^{2} \times ( { - 7n)}^{5} - {7n}^{5} \times {5m}^{5} - {7n}^{5} \times ( { - 7n)}^{5} = { - 25m}^{7} + {35m}^{5} {n}^{5} + {49n}^{10}$

2. 1)

${4}^{2} {c}^{2} - {3}^{2} = {(4c)}^{2} - {3}^{2} = (4c - 3) \times (4c + 3)$

2)

${4y}^{8} - {25y}^{12} = {y}^{8} \times (4 - {25y}^{4} ) = {y}^{8} \times (2 - {5y}^{2} ) \times (2 + {5y}^{2} )$

3)

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

3. 1)

${5x}^{2} + 10x - x - 2 + 3( {x}^{2} - 16) = 2( {4x}^{2} + 12x + 9) - 8 \\ {8x}^{2} + 9x - 50 = {8x}^{2} + 24x + 10 \\ 9x - 50 = 4x + 10 \\ - 15x = 60 \\ x = - 4$

2)

$(7x - 6 - 9) \times (7x - 6 + 9) = 0 \\ 7x - 15 = 0 \\ 7x + 3 = 0 \\ x = \frac{15}{7} \\ x = - \frac{3}{7}$

3)

${36c}^{2} - 48c + 16 - (16 {c}^{2} + 24c + 9) = 0 \\ {20c}^{2} - 72c + 7 = 0 \\ 2c \times (10c - 1) - 7(10c - 1) = 0 \\ 10c - 1 = 0 \\ 2c - 7 = 0 \\ c = \frac{1}{10} \\ c = \frac{7}{2}$

4)

${x}^{2} - 6x + 9 + {x}^{2} + 4x + 4 = (6 - 2x) \times (x + 2) \\ {2x}^{2} - 2x + 13 = 2x + 12 - {2x}^{2} \\ {4x}^{2} - 4x + 1 = 0 \\ ( {2x - 1)}^{2} = 0 \\ 2x - 1 = 0 \\ x = \frac{1}{2}$

5.

$x = \frac{6 + - \sqrt{( { - 6)}^{2} - 4 \times 1 \times 13 } }{2 \times 1} \\ x = \frac{6 + - \sqrt{36 - 52} }{2} \\ x = \frac{6 + - \sqrt{ - 16} }{2}$