Предмет: Математика, автор: elizavetastepanov

задание по алгебре за 7 класс, помогите пожалуйста

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Ответы

Автор ответа: сок111213

$(6a - \frac{1}{3} b)(6a + \frac{1}{3} b) = (6a) {}^{2} - ( \frac{1}{ 3} b) {}^{2} = 36 {a}^{2} - \frac{1}{9} {b}^{2}$

$(4x - \frac{1}{2} y) {}^{2} = (4x) {}^{2} - 2 \times 4x \times \frac{1}{2} y + ( \frac{1}{2} y) {}^{2} = 16 {x}^{2} - 4xy + \frac{1}{4} y {}^{2} = 16 {x}^{2} - 4xy + 0.25 {y}^{2}$

$( \frac{2}{3} y - 0.2)( \frac{2}{3} y + 0.2) = ( \frac{2}{3} y) {}^{2} - 0.2 {}^{2} = \frac{4}{9} {y}^{2} - 0.04$

$(0.3a - 2c)(0.3a + 2c) = (0.3a) {}^{2} - (2c) {}^{2} = 0.09 {a}^{2} - 4 {c}^{2}$

$(ax + 1.4) {}^{2} = {a}^{2} {x}^{2} + 2.8ax + 1.96$

$(2x - y) {}^{3} = (2x) {}^{3} - 3 \times (2x) {}^{2} \times y + 3 \times 2x {y}^{2} - {y}^{3} = 8 {x}^{3 } - 3 \times 4 {x}^{2} y + 6x {y}^{2} - y {}^{3} = 8 {x}^{3} - 12 {x}^{2} y + 6x {y}^{2} - {y}^{3}$

$(0.3a + 4b) {}^{3} = (0.3a) {}^{3} + 3 \times (0.3a) {}^{2} \times 4b + 3 \times 0.3a \times (4b) {}^{2} + (4b) {}^{3} = 0.027 {a}^{3} + 12b \times 0.09 {a}^{2} + 0.9a \times 16 {b}^{2} + 64 {b}^{3} = 0.027 {a}^{3} + 1.08a {}^{2} b + 14.4a {b}^{2} + 64 {b}^{3}$

$4x {}^{2} - 36 = 4( {x}^{2} - 9) = 4(x - 3)(x + 3)$

$x {}^{8} - y {}^{10} = ({x}^{4} ) {}^{2} - (y {}^{5} ) {}^{2} = ( {x}^{4} - {y}^{5} )( {x}^{4} + {y}^{5} )$

$0.3x + 1.2 {x}^{3} + 1.2 {x}^{2} = 0.3x(1 + 4 {x}^{2} + 4x) = 0.3x(2x + 1) {}^{2}$

$5 {a}^{2} - 10ab + 5b {}^{2} = 5( {a}^{2} - 2ab + b {}^{2} ) = 5(a - b) {}^{2}$

$0.125 {x}^{6} - 27 {y}^{3} = (0.5 {x}^{2} ) {}^{3} - (3y) {}^{3} = (0.5 {x}^{2} - 3y)((0.5 {x}^{2} ) {}^{2} + 0.5 {x}^{2} \times 3y + (3y) {}^{2} ) = (0.5 {x}^{2} - 3y)(0.25 {x}^{4} + 1.5 {x}^{2} y + 9 {y}^{2} )$

$64 {y}^{6} + 0.008 {x}^{6} = (4y {}^{2} ) {}^{3} + (0.2x {}^{2} ) {}^{3} = (4 {y}^{2} + 0.2 {x}^{2} )((4y {}^{2} ) {}^{2} - 4 {y}^{2} \times 0.2 {x}^{2} + (0.2 {x}^{2} ) {}^{2} ) = (4 {y}^{2} + 0.2 {x}^{2} )(16y {}^{4} - 0.8 {x}^{2} {y}^{2} + 0.04 {x}^{4} )$