. 100 БАЛЛОВ !!!! срочно,помогите!

Решите задачу:

$1)\; \; \dfrac{x^2+7x+12}{x+4}=\dfrac{(x+3)(x+4)}{x+4}=x+3\\\\\\\dfrac{x^2+x-6}{x^2-2x-15}=\dfrac{(x-2)(x+3)}{(x-5)(x+3)}=\dfrac{x-2}{x-5}\\\\\\\dfrac{3x^2+5x-2}{x^2-3x-10}=\dfrac{3\, (x-\frac{1}{3})(x+2)}{(x-5)(x+2)}=\dfrac{3x-1}{x-5}\\\\\\\dfrac{x^3-5x^2-14x}{x^2-2x-8}=\dfrac{x\, (x+2)(x-7)}{(x+2)(x-4)}=\dfrac{x\, (x-7)}{x-4}$

$2)\; \; x=0,25\\\\\dfrac{x^2+9x+20}{2x^2+10x}-\dfrac{x^2-x-12}{2x^2+6x}=\dfrac{x^2+9x+20}{2x(x+5)}-\dfrac{x^2-x-12}{2x(x+3)}=\\\\\\=\dfrac{(x^2+9x+20)(x+3)-(x^2-x-12)(x+5)}{2x(x+5)(x+3)}=\\\\\\=\dfrac{x^3+3x^2+9x^2+27x+20x+60-(x^3+5x^2-x^2-5x-12x-60)}{2x(x+5)(x+3)}=\\\\\\=\dfrac{8x^2+64x+120}{2x(x+5)(x+3)}=\dfrac{0,5+16+120}{0,5\cdot 5,25\cdot 3,25}=\dfrac{136,5}{8,53125}=16$