Question

Помогите срочно
алгебра

Answer 1

Номер 1.

а)

$(5a + frac{1}{3} b) times (5a - frac{1}{3} b) = 25 {a}^{2} - frac{1}{9} {b}^{2}$

б)

$(3x + frac{1}{3} )^{2} = 9 {x}^{2} + 2x + frac{1}{9}$

в)

$(ab - cx) times (ab + cx) = {a}^{2} {b}^{2} - {c}^{2} {x}^{2}$

г)

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

д)

$(0.4a - 10c) times (0.4 a+ 10c) = {0.4}^{2} times {a}^{2} - 100 {c}^{2} = \ = frac{4}{25} {a}^{2} - 100 {c}^{2}$

е)

$(ax - 3) ^{2} = {a}^{2} {x}^{2} - 6ax + 9$

Номер 2.

1)

а)

$(2a - b)(2a + b) + {b}^{2} = 4 {a}^{2} - {b}^{2} + {b}^{2} = 4 {a}^{2}$

б)

${(x + 7)}^{2} - 10x = {x}^{2} + 4x + 49$

в)

$9 {x}^{2} - (c + 3x)(c - 3x) = 18 {x}^{2} - {c}^{2}$

г)

$5 {b}^{2} - (a - 2b)^{2} = {b}^{2} - {a}^{2} + 4ab$

2)

а)

$(a - c)(a + c) - {(a - 2c)}^{2} = - 5 {c}^{2} + 4ac$

б)

${(x + 3)}^{2} - {(x - 3)}^{2} = 12x$

в)

${(a + 3c)}^{2} - (b + 3c)(b - 3c) = {a}^{2} + 6ac + 18 {c}^{2} - {b}^{2}$

г)

${(x - 4y)}^{2} + {(x + 4y)}^{2} = - 16xy$

д)

$(x - 3)(x + 3) - (x + 8)(x - 8) = {x}^{2} - 9 - ( {x}^{2} - 64) = \ = {x}^{2} - 9 - {x}^{2} + 64 = 55$

е)

$(2a + 1) times (2a - 1) + (a - 7) times (a + 7) = 5 {a}^{2} - 50$