Question

Помогите пожалуйста срочно надо!!!!!!!​

Answer 1

1. Виконаите Множения

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

$в)(2x - 3y)(3x - 2y) = 2x \times 3x - 2x \times 2y - 3y \times 3x - 3y \times ( - 2y) = 6x {}^{2} - 4xy - 9xy + 6y {}^{2} = 6x {}^{2} - 13xy + 6y {}^{2}$

$г)(4a {}^{2} + a + 3)(a - 1) = 4a {}^{2} \times (a - 1) + a \times (a - 1) + 3(a - 1) = 4a {}^{3} - 4a {}^{2} + a {}^{2} - a + 3a - 3 = 4a {}^{3} - 3a {}^{2} + 2a - 3$

2. Розкладіть на множники

$a)x {}^{2} - xb = x \times (x - b)$

$б)4c {}^{4} - 6x {}^{2} c {}^{2} + 8c = 2c \times (2c {}^{3} - 3x {}^{2} c + 4) = 2c \times (2c {}^{3} - 3cx {}^{2} + 4)$

$в)x(y - 1) - 3(y - 1) = (y - 1) \times (y - 3)$

$г)x {}^{3} + x {}^{4} + 2x + 2 = x {}^{3} \times (1 + x) + 2(x + 1) = (1 + x) \times (x {}^{3} + 2)$

3. Розв' яжіть рівняння

$a)5y {}^{2} - y = 0$

$y \times (5y - 1) = 0$

$y = 0$

$5y - 1 = 0$

$y = 0 \\ y = \frac{1}{5}$

$y _{1} = 0,y_{2} = \frac{1}{5}$

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$б)(6x + 5) - 2x(6x + 5) = 0$

$6x + 5 - 2x \times (6x + 5) = 0$

$(6x + 5) \times (1 - 2x) = 0$

$6x + 5 = 0$

$1 - 2x = 0$

$x = - \frac{5}{6} \\ x = \frac{1}{2}$

$x _{1} = - \frac{5}{6} ,x _{2} = \frac{1}{2}$

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$в)(10x + 9)x = 8 - (1 - 5x)(2x + 3)$

$10x {}^{2} + 9x = 8 - (2x + 3 - 10x {}^{2} - 15x)$

$10x {}^{2} + 9x = 8 - ( - 13 + 3 - 10x {}^{2} )$

$10x{}^{2} + 9x = 8 + 13x - 3 + 10x {}^{2}$

$9x = 8 + 13x - 3$

$9x = 5 + 13x$

$9x - 13x = 5$

$- 4x = 5$

$x = - \frac{5}{4}$