Question

80 Баллов. Помогите решить!
1)(2x^2+3x)/(x+2) *(x^2-4)/(2x+3)
2)(2x-1)/(x^2-6x+9) *(x^2-3x)/(1-2x)

Answer 1

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

$\frac{2x - 1}{x { }^{2} - 6x + 9} \times \frac{x {}^{2} - 3x}{1 - 2x} = \frac{2x - 1}{(x - 3) {}^{2} } \times \frac{x \times (x - 3)}{ - (2x - 1)} = \frac{ - 1}{x - 3} \times x = - \frac{x}{x - 3}$