Question

Помогите, срочно, даю 50 баллов

Answer 1

Разложить на множители:

$x^2+5x-14 =\\=\left(x^2-2x\right)+\left(7x-14\right)=\\=x\left(x-2\right)+7\left(x-2\right)=\\=\left(x-2\right)\left(x+7\right)$

$5x^2+7x-24 =\\= \left(5x^2-8x\right)+\left(15x-24\right) =\\= x\left(5x-8\right)+3\left(5x-8\right) =\\= \left(5x-8\right)\left(x+3\right)$

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

Сократить дробь:

$\frac{x^2+7x+12}{x+4} = \frac{\left(x^2+3x\right)+\left(4x+12\right)}{x+4}=\\\\= \frac{x\left(x+3\right)+4\left(x+3\right)}{x+4} = \frac{\left(x+3\right)\left(x+4\right)}{x+4} = x+3$

$\frac{3x^2-16x+5}{x-5}= \frac{\left(3x^2-x\right)+\left(-15x+5\right)}{x-5} =\\\\\frac{x\left(3x-1\right)-5\left(3x-1\right)}{x-5} = \frac{\left(3x-1\right)\left(x-5\right)}{x-5} = 3x-1$

$\frac{3x^2+5x-2}{x^2-4} = \frac{\left(3x^2-x\right)+\left(6x-2\right)}{\left(x+2\right)\left(x-2\right)} =\\\\ \frac{x\left(3x-1\right)+2\left(3x-1\right)}{\left(x+2\right)\left(x-2\right)} = \frac{\left(3x-1\right)\left(x+2\right)}{\left(x+2\right)\left(x-2\right)} =\frac{3x-1}{x-2}$