Question

помогите плиз срочно !!!

Answer 1

1)
$\frac{5 {x}^{2} }{x + 5} - \frac{6 {x}^{2} - 3x - 30}{ x + 5} - \frac{5 - 2 {x}^{2} - 7x}{x + 5} = \frac{5 {x}^{2} - (6 {x}^{2} - 3x - 30) - (5 - 2 {x}^{2} - 7x)}{x + 5} = \frac{5 {x}^{2} - 6 {x}^{2} + 3x + 30 - 5 + 2 {x}^{2} + 7x }{x + 5} = \frac{ {x}^{2} + 10x + 25}{x + 5} = \frac{ {(x + 5)}^{2} }{x + 5} = x + 5$
2)
$\frac{7 {x}^{2} - 4x }{2x - 5} - \frac{ {x}^{2} + 3x + 30}{2x - 5} - \frac{2 {x}^{2} - 7x - 5 }{2x - 5} = \frac{7 {x}^{2} - 4x - ( {x}^{2} + 3x + 30) - (2 {x}^{2} - 7x - 5) }{2x - 5} = \frac{7 {x}^{2} - 4x - {x}^{2} - 3x - 30 - 2 {x}^{2} + 7x + 5 }{2x - 5} = \frac{4 {x}^{2} - 25 }{2x - 5} = \frac{(2x - 5)(2x + 5)}{2x - 5} = 2x + 5$

Answer 2

1.

$\frac{5x^2}{x+5} -\frac{6x^2-3x-30}{x+5} -\frac{5-2x^2-7x}{x+5} =\frac{5x^2-6x^2+3x+30-5+2x^2+7x}{x+5} =\frac{x^2+10x+25}{x+5}=\frac{(x+5)^2}{x+5} =x+5$ (при х≠-5)

2.

$\frac{7x^2-4x}{2x-5} -\frac{x^2+3x+30}{2x-5} -\frac{2x^2-7x-5}{2x-5} =\frac{7x^2-4x-x^2-3x-30-2x^2+7x+5}{2x-5} =\frac{4x^2-25}{2x-5}=\frac{(2x)^2-5^2}{2x-5} =\frac{(2x-5)(2x+5)}{2x-5} =2x+5$ (при х≠2,5)